ATHEMATICS
FOR
HNICAL SCHOOI
WARk N A
( ) K
i Kh C < J I' ° CL
MATHEMATICS
FOR
TECHNICAL SCHOOLS
J. M. WARREN, B.A.,
Assistant Principal Day Schools
Central Technical School, Toronto
W. H. RUTHERFORD, MA., D.Paed.,
Director Department of Mathematics
Central Technical School, Toronto
TORONTO
THE COPP CLARK COMPANY, LIMITED
1921
Copyright, Canada, J921, by The Copp Clark Company, Limited,
Toronto, Ontario
PREFACE
In this book an attempt has been made to present the
subject of Elementary Mathematics in a way suitable to
industrial students in our technical schools. While it would
be manifestly impossible to deal with the mathematics of all
the industries in a book of this nature, yet we hope that
the fundamentals as herein presented will form a basis for a
wide range of industries.
No doubt experts in the various departments will have
suggestions to make as to how the book might be improved.
We will be very glad to hear from them in this connection.
With respect to the general plan of the work, we are
indebted to Dr. F. W. Merchant, Director of Technical
Education for Ontario, and Dr. A. C. McKay, Principal of
the Toronto Technical Schools. Thanks are due in a special
sense to Volney A. Ray, M.A., of the Department of Shopwork
in the Central Technical School in connection with the
chapter on " Mathematics of the Machine Shop," and to A. J.
Stringer, M.S.A., of the Department of Architecture and
Design in connection with the chapter on '• Application of
the Measures to the Trades." The cuts of Quick Change
Gears are by courtesy of the R. K. Le Blond Machine Tool
Company, Cincinnati, and those of the Planimeters by courtesy
of the Hughes Owens Company, Montreal. The drawings
were made by James Hanes a former student of our school.
June, 1921.
CONTENTS
Chapter Page
i. — The Fundamental Operations of Arithmetic . . 1
ii.— Fractions — Percentage 18
hi. — Weights and Measures — Specific Gravity . . 37
iv.— Square Root 50
v. — Application of Measures to the Trades . . .54
vi. — Algebraic Notation 71
vii. — Simple Equations 84
viii. — The Fundamental Operations of Algebra . . 92
ix. — Formulas 103
x. — Mensuration of Areas 108
xi.— Ratio and Proportion 134
xii. — Simultaneous Equations — Formulas (continual) . 140
xiii.— Graphs 150
xiv. — Mathematics of The Machine Shop .... 171
xv. — Logarithms 228
xvl— Mensuration of Solids ....... 241
xvii.— Resolution into Factors 263
xviii. — Indices and Surds 270
xix.— Quadratic Equations 27!)
xx.— Variation ' . . . 288
«
xxi. — Geometrical Progression 297
Miscellaneous Exercises 301
Tables— Decimal Equivalents. Weight and Specific
Gravity, Logarithms, Amttlogartthms . . . 315
Answers 323
Index 335
CHAPTER I.
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC.
1. The Symbols of Arithmetic are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0.
These symbols are called numbers, digits or figures. Their
values depend on how they are written with respect to each
other. When used separately or with commas between them
as above they denote one, two, three, four, five, six, seven,
eight, nine, zero. When written one after the other with no
marks between, their values are determined by their positions.
The established method of numeration, the Decimal System
(from the Latin word decern, ten) is based on the number ten.
For example 534 is read five hundred and thirty four. The
figure 4 being in the first place counting from the right indi
cates 4 units, the figure 3 being in the second place from the
right indicates ten times three units or thirty, the figure 5
being in the third place from the right indicates one hundred
times five units or five hundred. The following table indi
cates the values of the figures owing to their positions:
CD
a
S co *s
X3 G cd G
CD co XI *> co
3 T3 co *j T3 3
S 1 M S 1 , I } J
1 •§ H § •§ m ■ m £ 3 I P
S fl a o 5 c i
§ B H H W H P E
7 2 5 6438
in which a point called the decimal point is used to separate the
units figure from one having one tenth the value. Thus
7256438 is read seven thousand', two hundred and fifty six
and four hundred and thirty eight thousandths. The figures
following the decimal point are read as thousandths because
1
*J
T3
T3
a
o
s,
c3
CO
S3
a
O
3
xi
w
H
1.
36
2.
•734
3.
43689
4.
718965
9.
•34
10.
•435
11.
•03
12.
•075
2 MATHEMATICS FOR T?]CHXICAL SCHOOLS
the last figure is in the thousandths place. Thus • 13 would be
read thirteen hundredths because the last figure is in the
hundredths place. A whole number may be written with a
decimal point to the right of the units place.
Exercises I.
Write in words the following numbers:
5. 934
6. 73245
7. 43 124
8. 79861583
When the meanings of the figures in their relation to the
decimal point have been fixed, the figures to the right of the
decimal point are not read as above.
For example 13456 is read one hundred and thirty four
decimal five six or more generally one hundred and thirty
four point five six, that is the figures to the right of the decimal
point are merely named in their order going from left to right.
Exercises II.
1. Read the numbers in Exercises I making use of this
notation.
Express the following numbers in figures:
2. Four hundred and thirty four.
3. Seven hundred and forty eight and twenty six hun
dredths.
4. Six thousand, four hundred and eighty two and seven
tenths.
5. Five million, three hundred and nine thousand five
hundred and six and one hundred and twenty five
thousandths.
6. Five onethousandths.
7. Sixty five tenthousandths.
8. Three hundred and twenty five onethousandths.
9. Four hundred and seventy eight point three four.
10. Five thousand, three hundred and fifty point seven
eight six.
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 3
2. The Four Fundamental Operations. All computations
in Arithmetic are made by means of the four operations:
Addition or finding the sum.
Subtraction or finding the difference.
Multiplication or finding the product.
Division or finding the quotient.
3. Addition. The sign for addition is + (plus). Thus
6+4 means that 6 and 4 are to be added. The result
is called the sum. If we wish to add 6 ft. 4 in. and 3 ft. 2 in.
we must add in. to in. and ft. to ft. In a similar way when
adding numbers it is necessary to place tens under tens,
units under units, tenths under tenths and so on. In the
case of numbers having no decimal part this may be done by
keeping the margin on the righthand side in a straight line
and, in the case of numbers having decimal parts, by keeping
the decimal points in a vertical column.
For example to add 9, 75, 18, 324, 9678, 27436, and also
375, 12469, 75, 0023, 346058, 27, the arrangement is as
follows:
9 375
75 12469
18 75
324 0023
9678 346 058
27436 27
37540 536 0003
Each column is added beginning at the right. The sum
of the figures in the units column of the first case is 40, the
is placed in the units column, and the 4 is carried and added
to the figures in the second column since 40 units is equal to
4 tens and units. The sum of the figures in the tens column
with the 4 carried over is 24, the 4 is placed in the tens column
and the 2 is carried to the hundreds column and so on. In
a similar way beginning at the right the sum in the second
case is found.
1.
7.
10.
13.
MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises III.
Copy in your work book and add the following:
743
2.
1975
3.
1374
1589
4386
9281
642
721
4962
7593
15935
758
846
420
63
2572
5.
3521
6.
32842
13601
13635
73684
2354
2348
3943
728
7862
10026
19971
9143
70285
5392
8.
11864
9.
32125
1681
40621
7684
425 '
3259
135241
•85
7684
•13
324
23135
47002
»
13926
11.
21985
12.
2635
43584
43654
18923
936815
3985216
29712
72002
798005
43 • 002
73254
•792
1986
•006
43841
86812
13021
983521
12534
4798058
7648005
9875 1346
$ 8925
14.
si 728 36
15.
S32051
12163
25693
19281
287
2487
56853
1342
3425
40296
82978
17698
76834
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 5
4. Subtraction. The sign for subtraction is — (minus).
Thus 6 — 4 means that 4 is to be subtracted from 6. The
result is called the difference. In subtraction the numbers
are arranged as in addition, that is units under units, tens
under tens, and so on. For example, to subtract 872 from
2625 the arrangement is as follows:
2625
872
1753
2 is taken from 5 leaving 3. Since 7 cannot be taken from 2,
1 hundred or 10 tens is borrowed from 6 hundreds and 7 tens
are then subtracted from 12 tens leaving 5. In the third
column there are now only 5 hundreds in the upper line.
Since 8 cannot be subtracted from 5, 1 thousand is borrowed
from the 2 thousands and 8 hundreds are then subtracted
from 15 hundreds leaving 7 hundreds. The operation may be
performed by adding to the lower line instead of subtracting
from the upper line, thus 2 from 5 leaves 3, 7 from 12 leaves
5, 9 from 16 leaves 7, 1 from 2 leaves 1.
Exercises IV.
Copy the following examples in your work book and subtract :
1.
7963428.
6.
114238216.
2.
748512.
7.
2354813218.
3.
436271253.
8.
24150034.
4.
$168 45 $29 25.
9.
210539821.
5.
3464823921.
10.
638215421006
5. Multiplication. The sign for multiplication is X (mul
tiplied by). Thus 6X4 means that 6 is to be multiplied by 4.
The number multiplied is called the multiplicand, the number
by which it is multiplied is called the multiplier, the result
is called the product. Before the operation of multiplication
can be performed it is necessary to commit to memory the
multiplication tables following:
MATHEMATICS FOR TECHNICAL SCHOOLS
Multiplication Tables.
1
2
3
4
5
6
7
8
9
10
2
4
6
8
10
12
14
16
24
18
27
20
3
6
9
12
15
18
21
30
4
8
12
16
20
24
28
32
36
40
5
10
15
20
25
30
35
40
48
45
54
50
6
12
18
24
30
36
42
60
7
14
21
28
35
42
49
56
63
70
8
16
24
32
40
48
56
64
72
80
9
18
27
36
45
54
63*
72
81
90
10
20
30
40
50
60
70
80
88
90
100
11
22
33
44
55
66
77
99
110
12
24
36
48
60
72
84
96
108
120
In the table the second column gives the products when 1
is multiplied by 2, 2 by 2, 3 by 2 and so on to 12 by 2; the
third column gives the products when 1 is multiplied by 3,
2 by 3 and so on to 12 by 3. Similarly the seventh column
gives the products when 1, 2, 3 and so on up to 12 are mul
tiplied by 7.
To multiply 8345 by 7 the arrangement is as follows : 8345
7
58415
5X7 is 35 that is 3 tens and 5 units. The 5 is placed in the
units column and the 3 is carried to the tens column. 4 X 7 is 28
and when the 3 carried over is added the result is 31 tens the 1 is
placed in the tens column and the 3 is carried to the hundreds
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 7
column. 3X7 is 21 and when the 3 carried over is added
the result is 24 hundreds. The 4 is placed in the hundreds
column and the 2 is carried to the thousands column. 8X7 is
56 and with the 2 carried over the result is 58 thousands.
The 8 is placed in the thousands column and the 5 in the ten
thousands column.
6. Powers of 10. 10X10=100 and may be written 10 2 .
10X10X10=1000 and may be written 10 3 . 10 2 may be
called the second power of 10, 10 3 the third power and so on,
the figure placed to the right and above the ten being called
the index or exponent of the power. It may be observed that
the number of ciphers is the same as the index of the power.
7. Multiplication by 10 and its Powers. 436X100 = 43600.
72526 X 10 = 72526 since 6 hundredths multiplied by 10
becomes 60 hundredths or 6 tenths, 2 tenths multiplied by 10
becomes 20 tenths or 2 units and so on. Also 43 568X100 =
4356.800. The rule may be stated as follows: — To multiply
by 10 or its powers write the number with decimal point moved
as many places to the right as the number of ciphers in the power,
that is as many places as the index. Since 400 = 4X100 it is
evident that the product when multiplying by 400 may be
obtained by multiplying by 4 and then moving the decimal
point two places to the right.
Exercises V.
Copy in your work book the following examples and find
the products:
1. 173X9. 9. 7864X100. 17. 23 01X800.
2. 187X7. 10. 1475X8. 18. 00078X10 5 .
3. 769X8. 11. 298X6. 19. 00846X9000.
4. 34X7X4. 12. 298X60. 20. 1475X800.
5. 769X8X9. 13. 7864X10 2 . 21. 236896X10 6 .
6. 296X10. 14. 0067X7. 22. 6345X4X10 3 .
7. 345X100. 15. 0067X700. 23. 1273X8X10 4 .
8. 7645X10. 16. 42905X10 3 . 24. 765X10 3 X9.
8 MATHEMATICS FOR TECHNICAL SCHOOLS
8. When the multiplier contains more than one digit the
arrangement is as follows: 364X28= 364
28
2912
728
10192
364 is multiplied by 8 as before. When multiplying by 2
proceed as before but since the 2 is 2 tens the first figure 8
of the partial product is placed in the tens column and so on.
The partial products are added and the product 10192 obtained.
When the numbers have decimal parts as 13742X43 the
arrangement is as follows : 13 • 742
43
41226
54968
59 • 0906
The number of decimal places in the product is equal to the
total number of decimal places in the two numbers multiplied.
Exercises VI.
Copy the following examples in your work book and find
the products:
1. 364X9. 16. 054X721.
2. 793X8. 17. 829X431X08.
3. 436X11. 18. 7854X09X112.
4. 731X43. 19. 3 009X721 3X23 08.
5. 936X72. 20.. 543X 034X7 18.
6. 119X27. 21. • 035 X 728X436.
7. 4392X435. 22. 43 9 X 168 X 002.
8. 3854X729. 23. 143 • 5X7 25 X 075.
9. 9386X538. 24. 12961X324X503.
10. 1234X567. 25. 8421 X 158 X 072.
11. 2354X21. 26. 46X08X921.
12. 734183X36. 27. 736 X 98X4 12.
13. 9843X132. 28. 158 X 75 X • 1625.
14. 93 • 02 X 75. 29. • 0625 X 04 X 025.
15. 754X028. 30. • 1416X3 1416X35.
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 9
9. Division. The sign for division is •*■ (divided by).
Thus 3577 means that 35 is to be divided by 7, or that it is
required to find how many times 7 is contained in 35. The
number to be divided is called the dividend, the number by
which it is divided the divisor, the result of the division the
quotient. When a number is not contained an exact number
of times the part left over is called the remainder. Division
may also be indicated thus 8 7 5 .
10. Short Division. In general when the divisor is not too
large the method of short division is used.
Thus, 5852r7 = 7 /5852
836'
7 is contained in 58, 8 times and 2 to carry, 7 is contained in 25,
3 times and 4 to carry, 7 is contained in 42, 6 times. When
there is a remainder it is written over the divisor or reduced to
decimal form:
8/35826 8/3582600
or
4478 447825
11. Division by 10 and its Powers. To divide by 10 2 the
dividend may first be divided by 10 and the resulting quotient
then divided by 10. Since dividing by 10 makes each figure
equal to onetenth its original value owing to position, it is
evident that the result may be expressed thus: — to divide
by 10 or its powers move the decimal point as many places to
the left as the number of ciphers in the power of 10, that is as
the index of the power. Since 600 = 6 X 100 if 600 is the divisor
it is only necessary to divide by 6 and then move the decimal
point two places to the left. Hence the rule: — To divide by
a number ending with one or more ciphers move the decimal
point in the dividend as many places to the left as the number
of ciphers in the divisor and then divide by the part of the divisor
preceding the ciphers.
10 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises VII.
Copy in your work book the following examples and per
form the operations indicated:
1.
131948 h4.
7.
1325rlO 3 .
2.
2170944 r 12.
8.
W
3.
12348r7.
9.
12 • 5 r 500.
4.
W
10.
7659^90.
5.
•00632
11.
153 18 h 900,
4
12.
9/280765.
6.
176 MO 2 .
13.
7/32494.
12. Long Division. The method of long division is indicated
by the following example:
13/6942 /534
65
44
39
52
52
13 is contained in 69, 5 times. 13X5 is 65 which subtracted
from 69 leaves 4. Bring down 4 the next figure of the dividend .
13 is contained in 44, 3 times. 13X3 is 39 which subtracted
from 44 leaves 5. Bring down 2 the next figure of the dividend.
13 is contained in 52, 4 times. 13X4 is 52 which subtracted
from 52 leaves no remainder. When there is a remainder it
may be written over the divisor or changed to a decimal as
in short division 62563 ~ 39 = 62563 • 00 ■*■ 39
39/62563 / 1604^
. 39
235
234
163
156
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 11
or
39/62563 • 00 /160417 +
39
235
234
163
156
70
39
310
273
37
When the divisor or the dividend or both have decimal
figures the position of the decimal point in the quotient may
he obtained by paying attention to the following rules:
1. When the number of decimal places in the dividend exceeds
the number in the divisor, divide as if the divisor contained no
decimals and point off a number of decimal places in the quotient
equal to the number in the dividend minus the number in the divisor.
2. When the number of decimal places in the dividend is
less than the number in the divisor, annex zeros to the right of
the dividend until a sufficient number of decimals has been
obtained and proceed as before.
679/5720575/8425
54
32
2
885
2
716
1697
1358
3395
3395
12 MATHEMATICS FOR TECHNICAL SCHOOLS
Since there are 5 places in the dividend and 2 in the divisor,
the number in the quotient is 52 or 3 and the quotient is
therefore 8 425.
3430/16807/4
13 720
3 087
and the quotient is 004 since there are three decimal places
in the dividend and none in the divisor. If it is required to
carry the division to another decimal place add to the right
of the decimal and then divide into 30870 thus:
3430/ 168070/49
13 720 '
3 0870
and the quotient is 0049.
The following examples show a method often used in deter
mining the position of the decimal point.
Example:— Divide 433652 by 163.
2660
163/433 • 652
326
1076
978
985
978
72
Explanation: — When the divisor is an integer the point
in the quotient should be placed directly above the point
in the dividend and the division performed as in whole
numbers.
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 13
.e:— Divide 27'
4289 by 124.
22 12
12
4/27 4289/
24 8
262
248
148
124
249
248
1
Explanation: — When the divisor contains decimal figures move
the point in both divisor and dividend as many places to the
right as there are decimal places in the divisor. This is equiva
lent to multiplying both divisor and dividend by the same num
ber, 100 in the above, and does not change the quotient.
Then place the decimal in the quotient above the position
of the decimal point in the dividend and divide as in
whole numbers.
Exercises VIII.
Find results to two decimal places:
1.
5462 : 84.
4
7.
12354406.
2.
1024 f 16.
8.
73814926.
3.
31264 r 46.
9.
1934 43 4 136 3
4.
746215^352.
10.
138424034.
5.
8344621.
11.
1289427 4327
6.
7342 r 26 4.
12.
432198.
418.
13. Relative Importance of Signs of Operation. If only +
and — signs occur they may be operated in any order. Thus
12+3 — 2+9 — 6 = 16 in whatever order the signs are used.
14 MATHEMATICS FOR TECHNICAL SCHOOLS
If only X and ■*■ signs occur they must be operated in the
order given 1273X5 7 2 means that 12 is divided by 3, the
quotient multiplied by 5 and the resulting product divided
by 2.
If + and — signs occur together with X and 5 signs the
X and ■+■ signs must be used first and then the f and — signs
may be used in any order. Thus 1273+8X2672+7 =
4 + 163 + 7 = 24.
If brackets are used as in 36 t (4+8) the part within the
bracket is to be regarded as one quantity and the operation
would be 36712 = 3.
Exercises IX.
Find the values of:
1. 16r8+4X2X316X2^4.
2. 6025^5 + 1510074X2.
3. 17X3+27 7340X2H5.
4. 864 r 12 124 =31 +54 T 27.
5. 13X9X62+44 V4 17X22.
6. 4963V7 + 144T7214X9.
7. 1728s (362X12) + (13 X 12) s(8s 2).
14. Factors — Cancellation. The factors of the number are
the integers (meaning whole numbers) which multiplied
together give the number. Thus 3 and 5 are the factors of
15 since 3X5 = 15.
A number that has no factors but itself and unity (or 1)
is called a prime number. If a prime number is used as a
factor it is called a prime factor. Thus 2 and 5 are prime
factors of 20. When the same number is a factor of two or
more numbers it is said to be a common factor of those numbers.
Thus 3 is a common factor of 27 and 36. By means of factors
it is often possible to shorten the work in division. In 183 + 15
since 3 is a factor common to 183 and 15 we can divide by
it and then 183 t 15 = 61 t 5 = 12£.
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 15
This method, cancellation, may be used in finding the value
of such an expression as :
2 1 2
4X3X14X32 = 4X3XHX32 = 128 _ H ,
3X2X3X21 ' 3X2X3X21 9 " 9 '
1 1 3
First the 3 below the line is divided into the 3 above the line
and since 373 = 1 the 3's are cancelled by each other and
l's are placed in their stead. Similarly 2 below the line cancels
2 in the 4 above the line; next since 7 is a common factor of 14
and 21 it is divided into 14 giving 2 and into 21 giving 3. When
all common factors are cancelled the remaining numbers are
128
multiplied together giving —~ = 14?,.
Exercises X.
Find the values of:
 57X119X16 g 765X92X11
* 17X12X19. ' 368X9X10.
20X56X12 3218X 006X34
' 21X10X18. * 17X1609X003.
77X100X18X14 42X36X48
25X11X49X16. " 12X7X18.
1200X515X70X100  192X168X44
5X35X103. ' 4X21X22.
114X1728X999 1024X729X36
96X270X33. ' 144X9X18.
9925 + 14X7 10 2 X86X0625
50T2X18. ' 25X43X2.
256054 + 125X414x76 72X125X39
17X27+32X401618. 13X12X10 2 .
Exercises XI.
Applied Problems.
1 . In an electrical shop there were three motors, one weighed
278 lb., another 380 lb., and the third 475 lb. What was the
total weight?
16 MATHEMATICS FOR TECHNICAL SCHOOLS
2. Three coal sheds contained respectively 63821b., 14728 lb.,
24725 lb. How many tons in all three?
3. Electric light wire was run around the four sides of two
rooms. If the first room was 18 ft. long and 12 ft. wide; the
second 20 ft. long and 13 ft. wide, what was the total length
of wire required? (Electric lights require two wires).
4. A reel of wire contained 6425 ft. If 3226 ft. were used
on a certain job, how many ft. remained on the reel?
5. A reel of wire contained 7280 ft. If 2348 ft. were used
on one house and 1425 ft. on another, how many ft. were used
on both? How many ft. were left on the reel?
6. In the coalbin at the school there were 48,720 lb. of coal
at the beginning of the week. On Monday 11600 lb. were
used; Tuesday 12350 lb.; Wednesday 10718 lb. On Thursday
24600 lb. were received and 11880 lb. used. How much coal
was used during these days? How much coal was there in the
bin on Friday morning?
7. A machinist sent in the following order for bolts: 15
bolts, 3 lb. each; 21 bolts, 2 lb. each; 14 bolts, 4 lb. each;
9 bolts, 3 lb. each; 11 bolts, 6 lb. each. What was the total
weight of the order?
8. A wiring job required the following labour: 3 men for 4
hours each; 6 men for 5 hours each; 8 men for 9 hours
each; 2 men for 15 hours each. Find the total number of
hours on the job?
9. A rod is 72 in. in length. How many pieces 5 in. in length
can be cut from it? Would there be a remainder?
10. An engine requires 90 lb. of coal per mile. How far
could it run on 8 tons?
11. If 4 dozen screws weigh one pound, how many case
containing 24 screws could be filled from 30 lb. of screws'.'
12. A train runs from Toronto to Penetang, a distance of
101 miles, in 4 hours. What is the average rate per hour?
13. The cost of construction of a railway from Toronto to
Montreal, a distance of 333 miles, was $3,425,625. What
was the average cost per mile?
14. How many gallons of water would be discharged in an
hour by two pipes, if one discharged 18 gallons per minute and
the other 4 gallons more per minute?
THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 17
15. If 18 men working 8 hours a day, can do a piece of
work in 12 days, how many days will it take 24 men working
9 hours a day?
16. If a horseshoe weighs 8 oz., how many horseshoes will
36 lb. of steel produce. (1 lb. =16 oz.).
CHAPTER II.
FRACTIONS— PERCENTAGE.
15. Definition. A yard measure is divided, or marked off,
into three equal parts called feet so that:
1 foot = onethird (f) of a yard.
2 feet = twothirds (f) of a yard.
A foot rule is divided into twelve equal parts called inches
so that:
1 inch = Y2 of a foot.
5 inches = T \ of a foot.
9 inches = fe of a foot.
The symbols \, f, Y2, t \, T 9 y are called fractions because
they denote a part or fraction of something which has been
divided.
A fraction may then be defined as a number which denotes
one or more of the equal parts into which some thing or unit
has been divided. In the fraction f the number below the
line is called the denominator since it denotes or names the
parts into which the unit has been divided, the number above
the line is called the numerator, since it denotes the number
of parts taken. The numerator and the denominator are
called the terms of the fraction. A fraction expressed in this
notation is called a vulgar fraction.
Since 4 = 547 a fraction may also be regarded as a case of
indicated division.
16. Kinds of Fractions. When the numerator is less than
the denominator the fraction is said to be a proper fraction,
Ex. \, f, f. When the numerator is greater than the denomi
nator the fraction is said to be an improper fraction, Ex.
■f, I, \\. A combination of an integer (whole number) and a
fraction is called a mixed number, Ex. 3, 5&, 4.
18
FR A CTIONS— PERCENTAGE
19
17. To change a Fraction to another Equal in Value but
with Different Denominator.
p
§§
p
/ /yyyy//.
■p
m^V/Ws.
Wyyy
■
^%%
III
IIP
m
JfH
Fig. 1
In the above figure, in each case, we have a square If inches
to the side.
In the first case the shaded portion contains three of the
six equal parts and is onehalf the whole figure so that \=%.
In the second case the shaded portion contains eight of the
twelve equal parts and is twothirds of the whole figure so
that \=h
In the third case the shaded portion contains fifteen of the
twenty equal parts and is threefourths of the whole figure so
20 MATHEMATICS FOR TECHNICAL SCHOOLS
that f =ift. Further, § may be obtained from § by multiplying
numerator and denominator by the same number 3, also T \ may
be obtained from f by multiplying numerator and denominator
by the same number 4, and ^f may be obtained from f by multi
plying numerator and denominator by the same number 5.
From these illustrations it may be inferred that a fraction
is not changed in value when the numerator and the denominator
are multiplied by the same number.
If the above results are written f = \, ■& = §, B = i it may
be inferred that a fraction is not changed in value when the
numerator and the denominator are divided by the same number.
Exercises XII.
Change the following:
1. f to an equivalent fraction having 14 as denominator.
2. f to an equivalent fraction having 24 as denominator.
3. i\ to an equivalent fraction having 50 as denominator.
4. 4s to 7oths. 8. H to 144ths.
5. f to 40ths. 9. fc to 252nds.
6. 1 to 56ths. 10. ft to 189ths.
7. if to 108ths.
18. Reduction to Lowest Terms. A fraction is said to be in
its lowest terms when no number other than 1 will exactly
divide the numerator and denominator or, in other words, when
the numerator and denominator have no common factor.
The fraction  is in its lowest terms because 7 and 9 have
no common factor.
The fraction f is not in its lowest terms because 3 is a
common factor of 12 and 15 and dividing numerator and
denominator by the common factor, \l becomes f.
To reduce a fraction to its lowest terms divide both parts by
any common factor and continue the process until no further
division is possible. Ex. \\ '} = \ \ = f .
Exercises XIII.
Reduce to lowest terms:
1. . 3. 1. 5. 7. H&. 9. WV
2. A 4. £!£. 6. j 8. y 10. «*•
FRACTIONS— PERCENTAGE 21
19. To Reduce an Improper Fraction to a Mixed Number.
Example: — Reduce x  4 to a mixed number. 126 H 5 gives 25
for quotient and 1 for remainder and, as in division, may be
written 251. Therefore J f 6  = 25i. Hence the rule: — Divide
the numerator by the denominator and express as in division.
Note — any integer may be written in the form of a fraction
thus 25 = Y.
To Reduce a Mixed Number to an Improper Fraction.
Example: — Reduce 16f to an improper fraction.
Since in 1 there are 7 sevenths, in 16 there are 16X7 or 112
sevenths, and f is 3 sevenths, then 16f is 112+3 or 115
sevenths, therefore 16f= L p. Hence the rule: — Multiply
the whole number by the denominator and add the numerator
to the product. Take this result for the numerator and the
original denominator for the denominator.
Exercises XIV.
•Express the following improper fractions as whole or mixed
numbers:
1. f 4. V. 7.
2. J£. 5. V. 8.
3. V. 6. H 5  9.
Reduce to improper fractions:
15. 3. 17. 7 r \ 19.
16. 6J5. 18. llSaV 20.
20. Addition and Subtraction of Fractions. When fractions
have the same denominator they can be added by adding
the numerators, and subtracted by subtracting the numerators.
Exs. §+!=f. fif
When the denominators are not alike as \ and \ , they cannot
be added without first changing to equivalent fractions having
the same denominator.
1 1X2 o
w
10.
W
13.
HF
if
11.
25000
1 •
14.
.159
8 1 •
w
12.
w
ions:
121f.
21.
43If.
23.
722*
91f.
22
400011.
24.
392J
1 1X3
2 — 2X3
then,
14.13X25
2 1 3 ~ f, I — 6 >
also,
2_3_in_ i JL
3 5 15 la 15'
22 MATHEMATICS FOR TECHNICAL SCHOOLS
21. Least Common Multiple — Least Common Denominator
of Fractions. The fractions J, f , f , can be added only when the
denominators are alike. This may be any number of which the
different denominators are factors, but in practice it is customary
to take the smallest number containing the different denomi
nators. This number is then called the Least Common Multiple
(L.C.M.) of the denominators because it is the least number into
which the numbers will divide without remainder. It is also
called the Least Common Denominator (L.C.D.) of the fractions.
In the given case 12 is the L.C.M. of 2, 3, 4, then since 2 is
contained in 12, 6 times, the numerator and denominator
are multiplied by 6, so that \ =  % %, also § = T \, and f = T 9 T ,
men 2T3T4 nTntii  if
When the L.C.M. cannot beeasily determined by inspection
the following method may be used:
Find the least common multiple of 12, 14, 15, 16, 18, 20.
2/12 14 15 16 18 20
2/ 6
7 15
8
9 10
3/ 3
7 15
4
9 3
7 5 4 3
L.C.M. 2X2X3X7X5X4X3 = 5040.
Explanation: — Divide through by the least number which
is a divisor of two or more of the given numbers. Continue
this process until there is no number common to any two as a
factor. In the third line 3 and 5 are struck out because 15
is also in that line, and any number which is a multiple of 15
is also a multiple of 3 and 5. The L.C.M. is obtained as
indicated.
Exercises XV.
Find the values of the following:
i. i+f 6. a+aA.
11.
&2 I3 1 ^8'
2. *+f 7. f+^_ K 4._ ]V
12.
VI+4i2
3Ai 8. &+f++^.
13.
51+7^+6^
4 8"T3 — A 9. g +«°f ~f~3 2 ~~ 2
14.
VA + 7 9 .
5. !+&&. 10. 2i+4i+5£.
FRACTIONS— PERCENTAGE
Exercises XVI.
23
1. Four castings weigh respectively 8 lb., 5 lb., llf lb.,
and 7f lb. What is their total weight?
2. A piece of steel on a lathe is 1 in. in diameter. In the
first cut ^ in. is taken off, in the second cut A in., in the
third cut & in. Find the diameter of the finished piece.
*.£
+ H«^
7"
tf
Fig. 2
3. Find the overall length for the template in Figure 2.
r^hi
■ftH
Fig. 3
4. Find the missing dimensions in Figure 3.
5. A drawing calls for the following divisions:
3& in., 1\ in., 4fin., 8 in.
Find the overall dimensions.
TV
^■H*'^
;>r
Fig. 4
A crankpin has the dimensions given in Figure 4. If
in. is allowed at each end for finishing what must be the
6.
length of the rough forging?
Fig. 6
24
MATHEMATICS FOR TECHNICAL SCHOOLS
7. A drawing for a part of the end of a valve rod is given in
Figure 5. Find the missing dimension.
2»i*
Fig. 6
8. Find the missing dimensions, AB, CD, in Figure 6.
9. Find the missing dimensions x, y, z, in Figure 7.
10. Find the missing dimension in the upper part of the
height in Figure 7.
FRACTIONS— PERCENTAGE
25
22. Multiplication of Fractions — Consider the following
example: — A man left £ of his estate to his children, \ of this
being left to his eldest son. What fraction of the estate did
the eldest son receive ?
Fig. 8
We might represent this example by the above diagram.
ABCD represents the whole estate. The shaded part ABEF,
 of the whole, represents the part left to the children. Onehalf
of this is taken, BEHG, to represent the eldest son's share, i.e., \
of £ or \ Xf. We further observe that, of the eight squares
in the figure, the eldest son has three or § of the whole,
• Ivl3
• • 2^4 — 8
From this illustration we may infer that in order to mul
tiply \ by £ we multiply the numerators for a new numerator
and the denominators for a new denominator. Hence the
rule: — To multiply two or more fractions together, multiply
the numerators for a new numerator and the denominators
for a new denominator.
±IlUb, 4 Af — 4X7 — 2"8'
Frequently cancellation shortens the process.
Thus, fX4X£=W
Make a drawing to illustrate that ^ X f = 1% .
To multiply a mixed number by an integer one of two methods
may be used.
Thus to multiply 25 by 7. 25 X7 = 175
3 X i = \ = &%
175+2i = 177i
or, 25X7 = ^X7 = 5 P = 177i
26 MATHEMATICS FOR TECHNICAL SCHOOLS
To multiply two mixed numbers together change each to an
improper fraction and then multiply.
Thus to multiply 2\ by 43. 2§X4§ = JX¥ = W = 10iV
Exercises XVII.
Perform the operations indicated:
1.
1X4.
9.
tVXVX^.
17.
9iXfVX2.
2.
5X.
10.
1 3 V 7 V 54 y 9
1 8 AfffAif AtJ.
18.
31X^X1.
3.
3Xif
11.
T6 Xgj AXJATS
19.
ixvxii.
4.
ly3
2 A 8'
12.
4 y 5 1 V *
T3 A 3 A T o
20.
3X2X4.
5.
4 A iff.
13.
25 V 75 Y64
3 A "8" ^TO*
21.
13JXl^VX3i
0.
2 OI 8'
14.
15V 7 V 18
nAnA 2T
22.
1§X2X1 T V.
7.
l n f 5
4 0I 18'
15.
15^X4.
23.
"2 X3 T 9 X jg.
8.
3 V S V 2
Iff A a Aj.
16.
6iXf
24.
62 X4 T 3 ^X I3V
23. Division of Fractions. Consider Figure 8, page 25,
regarding multiplication of fractions.
ABCD represents the whole estate. The shaded part,
I of the whole, represents the part left to the children. This
is divided into two parts to represent the eldest son's share
i.e., f f 2 or I 7 \ . We observe that, of the eight squares in the
figure, the eldest son has three or f of the whole.
a • 2 _ 1
4 • 1 — 8 •
But in the previous illustration f = f X5.
• 2 . 2 _ 3 v/ 1
• • 4 • 1 — 4 A2
That is we may infer that to divide, f by \ we invert \
obtaining \ and then multiply f by \.
Hence the rule: — To divide one fraction by another invert
the divisor and proceed as in multiplication.
Thus to divide f by f . Invert the divisor f (i.e., write it f)
and multiply \ by f, .\ f s f = f Xf = if = T V
To divide a mixed number by a fraction change the mixed
number to an improper fraction and proceed as above.
Thus to divide 16 by . 16§^ = 4^ Xf = ^ = 81f.
FRACTIONS— PERCENTAGE 27
51
To reduce a complex fraction say = to a simple fraction
proceed as follows: 3
7  .23. " TT ' TT 2 A 23 **'
Exercises XVIII.
Find the results of the following:
1. f55. 7. i^J. 12. 3* 5 5* X^+^.
2. ff2. 8. 125***. 13 12HT3f+4fh4 T V
3A4. i_Xj 14. 5*Xtt3*2f
4. A3. y 'fxr 15 ljX2U9ft
5. tV"^"?' 10 8X3 "I" 6 031^1^
S. tt+f 11. 4fXl2iX T V 16 3X5*fif+liJ3*.
24. Decimal Fractions. The values of the figures in any
number depend upon their position with reference to the
decimal point.
Thus, 2 = 2 tenths = T %
• 25 = 2 tenths + 5 hundredths = T % + to = nnr
•342 = 3 tenths +4 hundredths +2 thousandths
— 3 4_4 4 2  342
~io 1 Toll 1 1000 — T00"ff'
In all such cases the decimal parts may be written as fractions
with some power of 10 as denominator, and are therefore
called decimal fractions.
25. To change a Decimal Fraction to an equivalent Vulgar
Fraction. It is evident that it is only necessary to write the
decimal, after removing the point, as numerator and 1 followed
by as many 0's as there are figures in the decimal as denominator.
Exercises XIX.
Change to equivalent fractions in their lowest terms:
1. 43. 6. 004.
2.
•04.
7.
•705.
3.
•752.
8.
• 1234.
4.
•7134.
9.
•016.
5.
•502.
10.
•000155.
28 MATHEMATICS FOR TECHNICAL SCHOOLS
26. To change a Vulgar Fraction to its equivalent Decimal
Fraction. Example: — Change  to its equivalent decimal.
1, . q _8 /l000 _ 19 ,
Example: — Change H to its equivalent decimal. H = 11 + 16 =
16/1 10000 / 6875
9 6
140
128
120
112
80
80
It is evident that, to change a fraction to its equivalent
decimal fraction, it is only necessary to perform the division
indicated after 0's have been placed to the right of the
decimal point.
Exercises XX.
Change the following fractions to their equivalent decimals:
1. b 4. H. 7. H 10. Hh
2. f 5. f. 8. «. 11. n
3 3 A 124 Q 24 10 1 27
27. Repeating Decimals. Example: — Change  to its equiva
lent decimal. ^ = 173 = 3 /1000
•3331
The division in this case would never end. £ therefore
produces what is known as a repeating decimal. This is
expressed by placing a period above the figure 3 .*.^= 3.
Example: — Change f to its equivalent decimal.
1 = 5^6 = 6/50000
•8333 +
In this case the decimal does not begin to repeat until the
second figure and is therefore called a mixed repeating decimal.
.'.  =83.
FRACTIONS— PERCENTAGE 29
The denominators in Exercises XX contain only 2's or 5's
or 2's and 5's as their factors. The fractions can be changed
into fractions having some power of 10 as denominators and
therefore give terminating decimals. All fractions such as
s, f, etc., having some factor other than 2 or 5 in the de
nominator, when expressed in their lowest terms, cannot be
changed into fractions having some power of 10 as denomi
nator and therefore give repeating or mixed repeating
decimals.
• Exercises XXI.
'Change the following to their equivalent decimals:
is 2i 3 1  4 2 ^ * fi 2 7 *
1. 9 . •£• 12 <>• T ** TF* ° T¥ ° TT 'T3"'
28. To change Repeating and Mixed Repeating Decimals to
their equivalent Fractions.
Example: — Change 24 to its equivalent fraction
• 24 = • 242424
100 times • 24 = 24 • 242424
1 times 21= 242424
Subtracting, 99 times 24 = 24
. 24 = i 4
That is to change a repeating decimal to its equivalent fraction
write the decimal, after removing the point, as numerator and
as denominator as many 9's as there are figures in the repeating
part.
Example : — Change • 34 to its equivalent fraction
•34= 34444
100 times 34 = 34444
10 times 34 = 3444
Subtracting, 90 times 34 = 31
. o\ _ 3 1
30 MATHEMATICS FOR TECHNICAL SCHOOLS
That is to change a mixed repeating decimal to its equivalent
fraction subtract the part which does not repeat from the whole
giving the numerator, and for denominator take as many 9's
as there are figures in the repeating part followed by as many
O's as there are figures which do not repeat.
Exercises XXII.
Express as fractions in their lowest terms:
1. 5 6. 369 11. 25306
2. 36 7. 3253 12.* 04726*
3. 36 8. 2516 13. 0036
4. 153 9. i42857 14. 0426
5. 369 10. 276
29. Percentage. The term "percent." usually written
%, is an abbreviation of the Latin "per centum" which means
by the hundred. Five percent. (5%) would be yf^ of the
quantity named. Percent, may be changed to a decimal
fraction.
Thus, 62% = rffr=62.
375% = fftf= : 375.
A decimal fraction of a quantity may be expressed as
percent.
Thus, 7 = T W = 70%
•89 = T Vxr = 89%
• 375 = Htf =375%
That is the decimal fraction may be changed to percent, by
moving the decimal point two places to the right. Also any
fraction may be changed to percent, by changing it to its equivalent
decimal fraction, and then moving the decimal point two places
to the right.
FRACTIONS— PERCENTAGE 31
Exercises XXIII.
1. In the following table supply the missing quantities:
%
Decimal
Fraction
Vulgar
Fraction
%
Decimal
Fraction
Vulgar
Fraction
%
Decimal
Fraction
Vulgar
Fraction
1
16f
100
•02
1
37
•25
•5
1
3
200
2
175
2*
6
386
•10
1
8
3
4
li'V
•9
'
350
2. Find 25% of 16, of 8, of 90, of 240.
3. 5 is what % of 10? of 20? of 40?
4. 8 is what % of 16? of 40? of 24?
5. What % of f is 2\1 27 of 600?
6. 20% of what number is 3? 7? 14? 17?
7. 68 is 15% less than what number?
8. 98 is 40% more than what number?
32 MATHEMATICS FOR TECHNICAL SCHOOLS
9. A gas bill was 25% higher last month than this. If it id
$6.46 this month how much was it last month?
10. How much water must be added to a 5% solution of a
certain liquid to make a 2% solution? (original solution 20
gallons) .
30. Short Methods. In practical work a large number of
decimal places is not needed. In all measurements the accuracy
depends upon the instruments, the methods used, and the thing
measured. It is only necessary that the error is small compared
with the quantity measured; a fraction of an inch in a dimension
of several feet would probably not make much difference.
In measuring to 001 inches it is not necessary to carry the
work to say 00001 inches. In any case of multiplication or
division it is only necessary to carry the result to one decimal
place more than the measurement. Thus if a measurement
of 7265 inches is multiplied by 3 1416 it is only necessary to
carry the work to four places of decimals, care being taken to
allow for numbers carried over from the fifth place.
Other short methods of multiplication and division may
be used.
To multiply by 5, 50, 500, etc., add 0, 00, 000, etc., to
the right of the number and divide by 2. Why?
To multiply by 25, 250, etc., add 00, 000 to the right of the
number and divide by 4. \\^hy?
To multiply by 125, add 000 to the right of the number
and divide by 8. Why?
To multiply by 33*, 16, 12$, 8, 6J. Add 00 to the
right of the number and divide by 3, 6, 8, 12, 16. Why?
By using the reverse process division by 33£, 16f, 12£,
125, etc., may be performed. Thus to divide by 33^ multiply
by 3 and divide by 100 or mark off two decimal places. Why?
To multiply a number ending in \ such as \Z\ by itself.
Multiply the number plus 1 by itself and add \ to the product.
Thus 13X13i = 14Xl3 + l.
FRACTIONS— PERCENTAGE 33
To multiply a number ending in 5 by itself, multiply the
number to the left of 5 by a number one greater than itself and
place 25 to the right of the number. Thus, 75X75, 7X8 = 56,
and the result is 5625.
Exercises XXIV.
Applied Problems.
1. From 2000 lb. of iron bars each weighing 80 lb. § is cut
up for bolts, £ for shafts and the remainder for studs. How
many bars are used for the different articles?
2. At 2^c. a pound, what will be the cost of 108 castings
each weighing 29 lb.?
3. An automobile runs at the average rate of 10 miles an
hour. How long will it take to go from Toronto to London,
a distance of 116 miles?
4. A  in. steel bar weighs 1914 lb. per foot. What will be
the cost of 5000 ft. of f in. steel bars if it cost $1.75 per 100 lb.?
5. Which is cheaper, and by how much, to have a 36^c.
an hour man take 12 hr. on a job or to have a 48c. an hour
man who can do the job in 9 hr.?
6. The weight of a foot of ^ in. steel bar is 106 lb. Find
the weight of a 20 ft. bar.
7. At 42^ c. an hr. what will be the pay for 21 \ days of 8
hours each?
8. If 2\ bundles of shingles are used on 82 sq. ft. of roof,
how many bundles will be used on 325 sq. ft. of roof?
9. How many pieces 5 in. long can be cut from a rod 27
ft. long?
10. A person spending ^, § and \ of his money has SI 19
left; how much had he at first?
11. If T 4 T of a house be worth $1969.92, what is the value
of ^ of the house?
12. Three men own a house worth $6250; one owns T 3 „ of it;
the second £ of it; what is the value of the third's share?
13. A man having 271 J acres of land, sold £ to one man
and  to another; what was the value of the remainder at
$323 • 68 an acre?
14. I want to mix up a pound of solder to consist of '4 parts
zinc, 2 parts tin and 1 part lead; what fraction of a pound of
each metal must I have?
34 MATHEMATICS FOR TECHNICAL SCHOOLS
15. An apprentice who is drilling and tapping a cylinder for
 in. studs, tries a f in. drill, but the tap binds, so he decides
to use a drill fa in. larger; what size drill will he use?
16. An 8 ft. bar of steel is cut up into 16 in. lengths; what
fraction of the whole bar is one of the pieces?
17. The time cards for a certain piece of work show 2 hours
and 15 minutes lathe work, 4 hours and 10 minutes milling,
2 hours and 20 minutes bench work; what is the total number
of hours charged to the job?
18. A gallon is about ^ of a cubic ft. If a cubic foot of
water weighs 62? lb., how much does a gallon of water weigh?
19. What is the cost of a casting weighing 432^ lb. at 6c.
a pound?
20. How many steel pins to finish 1 in. long can be cut
from an 8 ft. rod if we allow ^ in. to each pin for cutting off
and finishing?
21. A machinist whose rate is 675 cents per hour puts in a
full day of 8 hours and also 3 hours overtime. If he is paid
"time and a half" for overtime, how much should he be paid
altogether?
22. If an alloy is 67 copper and 33 zinc, how many pounds
of each metal would there be in a casting weighing 82 lb.?
23. A can do a piece of work in 25 days; B can do it in 30
days; C can do it in 35 days. In what time will they do it,
all working together?
24. A man earns $280 in 2\ months. If he spends in 4
months what he earns in 3 months, how much will he save in a
year?
25. From a farm of 125 T 3 7 acres there were sold at one time
2763 acres and at another 34 acres. How many acres
remained?
26. From an oil tank containing 375087 gallons there
leaked out each day 2f gallons. How many gallons remained
in the tank at the end of 25 days?
27. If the weight of a brass casting is approximately fifteen
and a half times that of its white pine pattern, what will be
the weight of a casting if the pattern weighs 15 oz.?
28. Since the shrinkage of brass castings is about \ in. in
10 in., what length would you make the pattern for a brass
collar which is required to be 6 in. long?
FRACTIONS— PERCENTAGE 35
29. How long will it take a drill making 134 revolutions per
minute (R.P.M.), at the rate of 012 in. per revolution, to
drill a hole 1^ in. deep?
30. A piece of wroughtiron 269 in. thick is to have two
H in. holes drilled through it. If the drill makes 112 R.P.M.,
what must be the feed to drill each hole in two minutes?
(The feed of a drill is the number of revolutions necessary to
cause the drill to descend 1 in.).
31. In drilling a bed plate a drill makes 67 R.P.M., and is
being fed to the work at the rate of 015 in. per revolution,
how deep will the hole be at the end of 4 minutes?
32. What will be the R.P.M. of a drill used for drilling a
lathe spindle 3024 in. long, the feed being 015 in. per revo
lution, and the time given to the job being 21 minutes?
33. What must be the R.P.M. of a drill, feeding at the
rate of 015 in. per revolution, to drill a hole 2\ in. deep in a
casting in 2 minutes?
34. A casting is to have a number of holes drilled in it 2\
in. deep with a highspeed drill making 260 R.P.M. What
must be the feed to drill each hole in f of a minute?
35. A man who owns f of a claim sold 6 of his share for
$2000. What decimal part of the claim does he still own and
what is the claim worth?
36. An engine rated at 1250 horsepower, is found to be
y o efficient. How many horsepower are available for
driving the machinery? How many are lost?
37. A board was cut into two pieces, one 8f in. and the
other 5^ in. long. If ^ in. be allowed for waste in cutting,
what was the length of the board?
38. A locomotive has a piston displacement of 12656 cu«
in. If the clearance space is 65% of the piston displacement,
what is the clearance space?
39. A merchant bought 15 carloads of apples of 212 barrels
each, 3 bushels in each barrel at 90c a bushel. He paid for
them in cloth at 25c. a yard. How many rolls of 477 yd. each
did he give?
40. A carload of pigiron weighs 90,000 lb. If 11$% of
this is used at once in the foundry, how much is left?
41. The diameter of two holes is 3 in. and the distance
between the sides of the holes is 3f in. What is the distance
from the outside of one hole to the outside of the other?
36 MATHEMATICS FOR TECHNICAL SCHOOLS
42. From a steel bar 27 f in. long were cut the following
pieces: — one 1\ in., one 6 in., one 3f in. long. If the length
of the bar was then 8 in., what was the amount of waste in
cutting?
43. A man, buying a house and lot, paid $2200 for the lot
and 62% more than that for the house. What did both
cost him?
44. A man invested $16,400 as follows: — 25% in an auto
mobile, 37^% in bank stock, and the remainder in an addition
to his house. How much did he invest in each?
45. An electrician has a reel of 300 ft. of copper wire. He
used at various times 50 ft., 32j ft., 109f ft. How much
wire was left? What percent, was left?
46. If § of the shell of a stationary boiler is considered as the
heating surface, how many square feet of heating surface are
there in a boiler containing 98^ sq. ft.?
47. A pump pumps 338 gallons to each stroke and the
pump makes 512 strokes per minute. How many gallons of
water will it pump per hour?
CHAPTER III.
WEIGHTS AND MEASURES— SPECIFIC GRAVITY.
31. Linear Measure. Linear Measure is used in measuring
lines and distance.
The fundamental unit of English Linear Measure is the
yard. It is the distance between two marks on a bronze
bar in the Royal Exchange, London, England.
Table.
12 inches (in.) = 1 foot (ft.).
3 ft. =1 yard (yd.).
5^ yd. = 1 rod.
320 rods = 1 mile.
Inches are commonly denoted by two strokes above the
figure. Feet are denoted by one stroke. Thus 6 in. is written
6" and 6 ft. is written 6'.
32. Surveyor's Measure. Surveyor's Measure is used in
measuring land.
Table.
7 92 in. = 1 link (li.).
100 li. = 1 chain (ch.).
80 ch. = 1 mile.
1 ch. =22 yd. =66 ft.
The chain in this table is known as Gunter's chain. It is
the one in general use for country surveys.
Engineers frequently use a chain, or steel tape, 100 ft. long.
The feet are usually divided into tenths instead of into inches.
33. Nautical Measure.
Table.
6 ft. =1 fathom.
120 fathoms = 1 cable.
6080 ft. = 1 nautical mile = 1 • 151 statute miles.
1 knot = a sailing rate of one nautical mile per
hour.
37
38 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises XXV.
1. How many yards in a mile?
2. How many feet in a mile?
3. One inch is what decimal of a yard?
4. One rod is what decimal of a mile?
5. Reduce 18 yd., 2 ft., 9 in. to inches.
6. Reduce 3 mi., 30 rods, 1^ yd. to feet.
7. Express 1 link as a decimal of a mile.
8. Express 1 in. as the decimal of a chain.
9. Change 4 chains, 15 links to links.
10. Change 26 yd., 1 ft., 2 in. to chains.
11. Change 4356 li. to feet.
12. Change 25 rods, 3 yd., 2 ft. to chains.
13. The world's record (Dec. 1919) for a destroyer was
45 • 5 knots. What is this in statute miles?
34. Metric Linear Measure. Metric is the adjective form
of the word metre which is a French word meaning "measure."
The earth's quadrant (one fourth of the circumference) was
measured by French engineers in 1799. One tenmillionth of
this length was taken as the length of the metre.
Table.
10 millimetres (mm.) = 1 centimetre (cm.)
10 cm. = 1 decimetre (dm.)
10 dm. = 1 metre (m.)
10 m. =1 decametre (Dm.)
10 Dm. = 1 hectometre (Hm.)
10 Hm. = 1 kilometre (Km.)
It may be seen that the prefixes have definite meanings:
milli = xoViT' centi = y^, deci = T x 7 , deca = 10, hecto = 100,
kilo = 1000.
35. Comparison of English and Metric Linear Measurements.
1 in. =25399 cm. (254 cm. approx.).
lcm. =3937 in.
lmile =160935 Km. (161 Km. approx.).
1 Km. =621 miles.
1 m. =393707 in. (3937 in. approx.).
Make calculations to test the accuracy of the above table.
WEIGHTS AND MEASURES— SPECIFIC GRAVITY
39
Exercises XXVI.
1. Measure the perimeter of the room with both metre stick
and yard stick. Make drawings to scale in your laboratory
book. Change the result in the English system to the Metric
system and compare.
2. Do the same as in 1 for the door, table, etc.
3. Write all the measurements in the Metric system in
terms of the metre.
4. Fill in the omitted entries in the following:
Unit
Equivalent
Inches
Feet
1 cm.
1 dm.
1 m.
1 Dm.
1 Hm.
1 Km.
5. A piece of steel bar is laid off to a length of 438 cm.
Find this length in feet and inches.
6. The thickness of a steel plate is f ". Find the thickness
in cm. and dm.
7. A speed of 200 ft. per second is how many Km. per
second?
40 MATHEMATICS FOR TECHNICAL SCHOOLS
8. When a body falls freely from rest it increases in speed
each second 322 ft. per second. Express this in cm. per
second each second.
9. An express train is travelling at the rate of 50 miles per
hr. Express this in Km. per minute.
10. Find the difference in cm. between the lengths of two
steel rods, one of which is 48' long and the other 48" long.
Square Measure. In measuring areas or surfaces, the
inch, foot, yard, etc., can no longer be used.
It is necessary to use the square inch, the
square foot, the square yard, etc.
By a square inch is meant a surface one
inch long and one inch wide.
Thus in measuring surfaces two dimen
sions, length and breadth, are used.
Table.
144 square inches (sq. in.) = 1 square foot (sq. ft.).
9 sq. ft. =1 square yard (sq. yd.).
30j sq. yd. =1 square rod (sq. rod).
160 sq. rods = 1 acre.
10 sq. chains = 1 acre.
640 acres =1 square mile (sq. mi.).
Make drawings to scale in your laboratory book and illus
trate the truth of the first three lines in the above table.
37. Metric Square Measure.
Table.
100 square mm. (sq. mm.) = 1 square cm. (sq. cm.)
100 sq. cm. = 1 square dm. (sq. dm.)
100 sq. dm. = 1 square m. (sq. m.)
100 sq. m. =1 square Dm. (sq. Dm.)
100 sq. Dm. = 1 square Hm. (sq. Hm.)
100 sq. Hm. = 1 square Km. (sq. Km.)
Make drawings to scale in your laboratory book and illus
trate the truth of each line in the above table.
WEIGHTS AND MEASURES— SPECIFIC GRAVITY 41
38. Comparison of English and Metric Square Measure.
Table.
1 sq. in. =64516 sq. cm.
1 sq. cm. = • 155 sq. in.
1 sq. ft. =0929 sq. m.
lsq. m. =10764sq. ft.
1 sq. yd. = 8361 sq. m.
1 sq. m. =1 196 sq. yd.
Make calculations to test the accuracy of the above table.
Exercises XXVII.
1. Find the area of the floor of your classroom in square
metres and also in square feet. Make drawings to scale in
your laboratory book. Change the area in square metres to
square yards and compare.
2. Find the area of a page of your laboratory book in sq.
in. and also in sq. cm. Test as in preceding question.
3. Perform similar experiments by measuring the school
yard, the door, table, the teacher's desk, etc.
4. Change one acre to sq. yd.
5. Express 4 sq. rods, 25 sq. yd., 7 sq. ft., in sq. ft.
6. Express 5 sq. rods, 8 sq. yd., 5 sq. ft., as the decimal of
an acre.
7. Express 5 sq. yd., 3 sq. ft., 18 sq. in., as sq. in.
8. Express 4 sq. ft., 85 sq. in., as the decimal of a sq. yd.
9. A square field measures 20 rods to a side. Find its area
in acres.
10. A steel plate in the form of a rectangle is 18^" long by
6j" wide. Find the area in sq. ft.
11. A numberplate on an automobile is 21" long by b\"
wide. Find area in sq. ft.
12. A rectangular garden 2\ chains wide contains f of an
acre. How many feet long is it?
13. How many sq. ft. of glass are there in a box containing
72 panes each 12" by 16"?
14. How many sq. yd. are there in the walls of a room 15'
6" long, 12' wide, and 9' 4" high?
15. A rectangular piece of land measures 1200 links by 180
links. What is its area in acres?
42
MATHEMATICS FOR TECHNICAL SCHOOLS
16. How many bricks 8 in. long and 4 in. wide will pave a
yard 116' long and 46' wide?
17. Find the cost of laying a concrete walk 400 yd. long
and 4 ft. 8 in. wide at 60c. a sq. yd.
18. Find the cost of painting both sides of a tight board
fence 80' long, 5' 3" wide at 7c. a sq. yd.
19. How many boards each 12' long and 10" wide will
be required to build a fence 60 yd. long and 4 ft. high?
20. How many sq. ft. of tin will be necessary to line the inside
of an open box whose external measurements are 4' long, 3'
8" wide and 2' 10" deep, if the material in the box is 2" thick
and 10% is allowed for cutting and joining the tin?
39. Cubic Measure. In the measurement of surfaces in the
preceding sections two measure
ments, length and breadth, were
used. The areas resulting were ex
pressed in square inches, square
feet, etc.
If it is required to measure the
volume of solids, the dimensions,
length and breadth must be taken
into account and in addition
another dimension — thickness.
By a cubic inch is meant the
volume of a cube, 1 inch on each
edge, Figure 10.
Volumes of solids are measured in cubic inches, cubic feet,
cubic yards, etc. Table.
1728 cu. in. = 1 cu. ft.
27 cu. ft. = 1 cu. yd.
128 cu. ft. = 1 cord (8'X4'X4').
Make drawings to scale in your laboratory book and illus
trate the truth of each line in the above table.
40. Metric. Cubic Measure.
Table.
1000 cubic millimetres (c.mm.) = 1 cubic centimetre (c.c.)
1000 c.c. = 1 cubic decimetre (c.dm.)
1000 c.dm. = 1 cubic metre (cm.)
1000 cm. = 1 cubic decametre (c.Dm.)
1000 c.Dm. = 1 cubic hectometre (c.Hm.)
Fig. 10
WEIGHTS AND MEASURES— SPECIFIC GRAVITY 43
41. Comparison of English and Metric Cubic Measure.
Table.
1 cu. in. =16387064 c.c. =16387 c.c. (approx.)
1. c.c. =. 06102 cu. in = 061 cu. in. (approx.)
1 cu.ft. = 02831 cm. = 028 cm. (approx.)
1 cm. =353163 cu. ft. =35316 cu. ft. (approx.)
Make calculations to test the accuracy of the above relations.
Exercises XXVIII.
1. Find the volume of the top of the laboratory table in cu.
ft. and in cm. Make drawings in your laboratory book.
Change from one system to the other and compare.
2. Find the volumes of the various rectangular models in
the laboratory, in cu. in. and also in c.c. Make drawings in
your laboratory book. Change from one system to the other
and compare.
3. Change 8 cu. yd., 9 cu. ft., to cubic feet.
4. Change 3 cu. yd., 2 cu. ft., 8 cu. in., to cubic inches.
5. A rectangular vessel is 15" long, 6£" wide and 4" deep,
inside measurements. Find its volume in cubic centimetres.
6. A gravel bed whose surface has an area of 2 acres, con
tains gravel to a depth of 10". How many miles of road 12'
wide can be covered with the gravel if it be spread to a uniform
depth of 7"?
7. The outside measurements of a cubical box, with a lid,
are 3' 4" long, 2' 8" wide and 1' 10" deep. If the box is made of
1" material, how many cu. ft. of material are there in the box?
How many cubic metres will it hold?
8. A cubical cistern, without a lid, 4' 4" long, 4' 4" wide and
6' 8" deep, outside measurements, is made of plank 2" thick.
How many cu. ft. of material are there in the box? How many
cu. ft. of water will it hold?
9. A pile of wood 10' long, 4' wide and 6' high was sold for
$20.00. What was the price per cord?
10. A woodyard 20' long and 18' wide is filled with cord
wood to a height of 6'. What is the wood worth at $8.50 a
cord?
11. If 1 cu. yd. of earth make a load, how many loads will
be removed in excavating for a foundation 4' deep, 36' 3"
long, and 24' wide?
44 MATHEMATICS FOR TECHNICAL SCHOOLS
12. The end of a rectangular bar of iron is a square f " to
the side. How many c.c. are there in 4' of the bar?
13. In excavating a tunnel 374,166 cu. ft. of earth were
removed. If the length of the tunnel was 492 ft. and the
width 39 ft., what was the depth?
14. In making a tender for some excavating a contractor
notes that the excavation is in the shape of a rectangle 11'
wide, 86' long at the top and has a depth of 8'. What will it
cost him to excavate it at 40c. a cu. yd.? What must he bid
to make a profit of 15%?
15. Rain falling uniformly for 5 hours on a roof, whose
dimensions are 30' by 15', fills a tank 6' 3" by 3' by 2' 6". Find
the depth of the rainfall per. hour.
16. The ice on a pond whose area is £ of an acre is 10"
thick. How many cu. ft. of ice may be removed?
42. Measures of Weight. The fundamental unit of English
weight is the pound. There are the pound Avoirdupois and
the pound Troy.
The pound, Avoirdupois, is equal to the weight of 7000
grains (plump grains of wheat) and is used for all ordinary
purposes of weighing. The pound, Troy, is equal to 5760
grains and is used in weighing gold, silver and precious stones.
Table — Avoirdupois Weight.
16 drams =1 ounce (oz.).
16 oz. = 1 pound (lb.) =7000 grains.
100 lb. =1 hundredweight (cwt.).
20 cwt. = 1 ton.
2240 lb. = 1 long ton.
Table— Troy Weight.
24 grains = 1 penny weight (dwt.)
20 dwt. = 1 oz.
12 oz. =1 lb. =5760 grains.
43. Metric System of Weights. The fundamental unit of
metric weight is the kilogram which is the weight of 1 litre,
equal in volume to 1 cubic decimetre, of distilled water under
fixed conditions of temperature and pressure.
WEIGHTS AND MEASURES— SPECIFIC GRAVITY 45
Table.
10 milligrams = 1 centigram (eg.)
10 eg. = 1 decigram (dg.)
10 dg. = 1 gram (g.)
10 g =1 decagram (Dg.)
10 Dg. = 1 hectogram (Hg.)
10 Hg. = 1 kilogram (Kg.)
44. Comparison of English and Metric Systems of Weights.
Table.
1 gram =15432 grains.
1 ounce = 28 35 grams.
1 pound (avoirdupois) =453*6 grams.
= 4536 kilograms.
1 kilogram =22046 pounds.
1 metric ton = 1000 kilograms.
 =22046 pounds.
Knowing any one of the above relations test the accuracy
of the others.
45. Measures of Capacity. The fundamental unit of
capacity in the English system is the gallon, which contains 10
pounds of distilled water under fixed conditions of temperature
and pressure.
46. Liquid Measure — used in measuring liquids.
Table.
4 gills = 1 pint (pt.)
2 pt. =1 quart (qt.)
4 qt. =1 gallon (gal.)
47. Dry Measure — used in measuring grains, vegetables, etc.
Table.
2 pints = 1 quart (qt.)
4 qt. =1 gallon (gal.)
2 gal. = 1 peck (pk.)
4 pk. = 1 bushel (bu.)
46 MATHEMATICS FOR TECHNICAL SCHOOLS
48. Metric System. The fundamental unit of measure
ment is the litre and is equal in volume to one cubic decimetre.
Table.
10 millilitres = 1 centilitre (cl.)
10 cl. = 1 decilitre (dl.)
10 dl. = 1 litre (1.)
10 1. =1 decalitre (Dl.)
10 Dl. = 1 hectolitre (HI.)
10 HI. = 1 kilolitre (Kl.)
= 1 cu. metre
49. Comparison of Capacity Tables with Cubic Measure.
1 litre =61024 cu. in. (approx.)
= •22 gal.
1 gal. =454 1.
1 cu. ft. = 28 38 litres.
= 62321 gal.
277274 cu. in. = 1 gal. 231 cu. in. = 1 gal. (American).
50. Specific Gravity. The specific gravity (sp. gr.) of a
substance is its weight as compared with the weight of an equal
volume of pure water.
Since the weight of a fixed volume of water is known we
can find the weight of an equal volume of any substance if we
know the specific gravity.
Example: — Find the weight of 8 cu. ft. of steel if its sp. gr.
is 7.8.
Solution: — 1 cu. ft. water weighs 62321 lb.
1 cu. ft. steel weighs 62321X78 lb.
8 cu. ft. steel weighs 62321X78X8 lb.=
388883 lb.
Exercises XXIX.
1. How much space will be filled by 14 tons of wrought
iron (sp. gr. 77)?
2. Find the average sp. gr. of a piece of brick construction
weighing 114 lb. per cu. ft.
WEIGHTS AND MEASURES— SPECIFIC GRAVITY 47
3. If 13 litres of milk weigh 1339 kilograms, what is the
sp. gr. of milk?
4. A tunnel 625 yd. long having a crosssection of 64 sq.
yd. is excavated through rock of sp. gr. 27. Find the weight
of rock removed.
5. If 3 litres of alcohol weigh 237 kilograms, what is the
sp. gr. of alcohol?
51. Measure of Time:
Table.
60 seconds (") = 1 minute (1').
60 minutes = 1 hour.
24 hours = 1 day.
7 days = 1 week.
365 days, 5 hours, 48 minutes, 48 seconds = 1 year.
As the calendar year of 365 days is nearly 6 hoars less than
the above, correction is made as follows: — Every year whose
number is divisible by 4 is a leap year and contains 366 days,
the other years containing 365 days, except that the century
years are leap years only when the number of the year is
divisible by 400.
The year is divided into 12 months: — January (Jan.),
February (Feb.), March, April, May, June, July, August
(Aug.), September (Sept.), October (Oct.), November (Nov.),
December (Dec).
" Thirty days hath September, April, June and November."
The other months, except February, have 31 days each.
February has 29 days in leap years and 28 days in all other
years.
Exercises XXX.
1. Compute the actual number of days from Sept. 23, 1919,
to April 6, 1920.
2. A note bearing interest from March 8, 1899, was paid on
July 5, 1900. Compute the interest period.
3. Reduce to the lowest denomination named: — 4 weeks,
3 days, 15 hr. 23 min.
4. How many hours between 10 A.M. Jan. 1, 1920, and
6 P.M. March 3, 1920.
48 MATHEMATICS FOR TECHNICAL SCHOOLS
52. Miscellaneous Measures :
Counting Tables.
12 things =1 dozen (doz.).
12 doz. = 1 gross.
12 gross = 1 great gross.
20 units = 1 score.
Stationers' Tables.
24 sheets = 1 quire.
20 quires = 1 ream.
3 reams = 1 bundle.
5 bundles = 1 bale.
Exercises XXXI.
1 Calculate the volumes of a number of the rectangular
solid models in the laboratory and estimate their weights in
both systems. Change from one system to the other and check.
2. Fill in the omitted entries in the following:
Quantity
Volume
Weight
cu. in.
c.c.
wt. in lb.
wt.
in Kg.
2 pints water
3 qt. water
1 cu. ft. water
1 gal. water
277274
10
10 c.c. water
10
1. Kl. water
WEIGHTS AND MEASURES— SPECIFIC GRAVITY 49
3. A rectangular tank is 25 m. long, 14 m. wide, and
•98 dm. deep. Find its capacity in litres. Find the weight
of water it will hold in grams.
4. The thickness of a steel plate is f ". If the plate has an
area of 400 sq. dm., find its volume in cu. in. and its weight in
lb. if 1 cu. in. of steel weighs 283 lb.
5. A block of granite weighs 2\ tons. Find, its weight in
kilograms.
6. Find the weight in grams of the air in a room 16' X 10'
and 9' high, if the air is 00128 times as heavy as water.
7. Find the number of litres in a rectangular tank 8'X
6'6"X4'3".
8. How many gallons of water are contained in a tank 6
metres long, 34 metres wide, and 27 metres deep?
9. A concrete watering trough is Z\' wide, 8' long and 2'
deep outside while inside the basin is 2' 10" wide, 7' 4" long
and 1' 6" deep. What is its weight if a cu. ft. of concrete
weighs 145 lb.? If the concrete was mixed in the proportion
of 1 cement, 2 sand, 3 stone, and 1 cu. yd. dry material makes
1 cu. yd. concrete, how many bags of cement were used.
(1 bag = l cu. ft.)?
CHAPTER IV.
SQUARE ROOT.
53. The Square of a Number is the product obtained by
multiplying the number by itself. Thus the square of 5 =
5X5 = 25.
The square root of a given number is that number whose
square is the given number. Thus the square root of 25 is
5 because 5X5 = 25.
Square root is indicated by prefixing the symbol y/ to the
given number. Thus \A>4 denotes the square root of 64.
When a number is small the square root may be found by
inspection or by means of the factors of the number. Thus
1225 = 5 X5X7 X7 = 5 2 X7 2 so that V1225 =V(5 2 X 7 2 )
= 5X7 = 35.
The following general method may be used for finding the
square root. To find the square root of 13264164.
13264J64 /3642
9
66 426
396
724 3041
2896
7282 145 64
145 64
Explanation: — Beginning at the decimal point, separate the
number into groups of two figures each, counting both to the
right and the left. Find the greatest square in the lefthand
group and write its square root as the first figure of the root.
In the example, 9 is the greatest square in 13, and 3 is the
first figure in the root.
60
SQUARE ROOT 51
Subtract the square from the lefthand group and to the
remainder bring down the next period to the right, thus forming
a new dividend.
In the example, 9 is subtracted from 13 and along with the
remainder 4 the next group 26 is brought down, giving 426
as the new dividend. Divide the new dividend, with its
righthand figure omitted, by twice the part of the root already
obtained and annex the result to both the root and the divisor.
Multiply the complete divisor by the last figure of the
root obtained, subtract, and bring down the next group to
form a new dividend as before.
In the example the 3 in the root is doubled giving 6, 6 is
now divided into 42 giving 6, and this figure is placed to the
right of the 6 already in the divisor and also as the second
figure of the root. Although 6 divided into 42 gives 7, if this
result is taken the result 67X7 gives a quantity too great to
subtract from 426, so that 6 must be taken instead. Proceed
in this manner until all the groups are used.
For every group to the right of the decimal point there must
be a decimal figure in the root.
When the number is not an exact square the root may be
obtained to any number of decimal places.
Exercises XXXII.
Find the square root (correct to four decimal places) of:
1. 2025. 2. 39601. 3. 15129. 4. 106929. 5. 1369.
6. 3. 7. 12 186. 8. 1432041. 9. 5432. 10. 06285.
11. Find the length in yards of the side of a square 10 acre
field.
12. A square pipe has an area of 1360752 sq. in. What is
the length of its side?
13. An outlet on a heating system is 4' 4" wide and 18"
high. A pipe leading from it must have the same area and
must be square. Find the size of the square pipe.
14. Would it be cheaper to build the square pipe or one of
the same dimensions as the outlet? Why?
52
MATHEMATICS FOR TECHNICAL SCHOOLS
15. A steel plate is rectangular in shape, 18"X14". Find
the side of a square plate of the same area.
54. One of the most Valuable Practical Uses of Square Root
is in finding the third side of a rightangled triangle, when
two of its sides are given.
In the adjoining figure ABC
is a rightangled triangle with
the sides AB and BC 3 in.
and 4 in. respectively (to
scale) .
Squares are described on
the three sides and divided
into smaller squares as in
dicated. If we make tests
with dividers we will find
that the small squares are
equal throughout the figure.
We will also notice that the
number of small squares in
the square on AC is equal
to the total of the number of small squares in the squares
on AB and BC.
From this experiment we derive: — In a rightangled triangle
the square on the side opposite the right angle {hypotenuse) is
equal to the sum of the squares on the other two sides.
Exercises XXXIII.
1. Find the distance from corner to corner of a square
piece of tin which contains 100 sq. in.
2. A room is 40' X 28'. Find the length of a diagonal.
3. If the above room is 16' high, find the distance from
any corner to the diagonal corner of the ceiling.
4. A baseball diamond is in the form of a square 90' to the
side. Find the distance from "first" to "third."
5. A boy was flying a kite with a string 650' long. If the
distance, from where the boy was standing, to a point directly
under the kite was 450 ft. how high was the kite?

Fig. 11
SQUARE ROOT 53
6. A tree broke in such a way that the top struck the ground
30' from the base of the tree. What was the height of the
tree, the broken part being 60 ft. long?
7. A ladder, 42' long, placed with its foot 24' from a wall,
reached within 2' of the top. How near the wall must the
foot of the ladder be brought in order that it may reach the
top?
CHAPTER V. •
APPLICATION OF MEASURES TO THE TRADES.
55. Stone Work. In stone work it is difficult to get any
fixed method of estimating the cost. One job will have a
set of conditions which do not exist in another, hence the
contractor will make an allowance in one that he would not
regard as necessary in another. There are two kinds of stone
work, rubble or rough and ashlar or squared. In rubble
work the toise is the common unit of measurement. This is
used both in estimating the amount of stone required and in
the cost of the work.
Table.
10 tons rubble = 1 toise (approx.)
1 toise — in wall = 162 cu. ft. (approx.)
1 toise — measured loose = 216 cu. ft. (approx.)
In ashlar work the unit for estimating either the amount of
stone necessary, or the cost of laying, is the cubic foot. The
labour for dressing the stone is figured by the square foot. The
minimum thickness of stone work for facing is 4 in. increasing
in thickness as requirements demand.
Exercises XXXIV.
1. How many toise of rubble will be required for the founda
tion of a house 40' 0" X 32' 0". the stone work being 5' 0"
high and 18" thick?
2. A cellar is 23' 6" wide by 35' 8" long and 6' 6" high. If
the wall is 16" thick and has two openings each 3' 3" X 2' 3",
find the number of toise of stone required.
3. The basement walls for a house 26' 0" wide and 38' 0"
long are to have 6 windows each 3' 0" X 2' 0". The walls are
to be 7' 0" high and 18" thick, (a) Find the cost at $20.00
a toise if the actual volume be estimated and 5% be allowed for
extra work on openings. (6) Find the cost at $18.00 a toise if
corners be doubled and only 50% of the openings be deducted.
54
APPLICATION OF MEASURES TO THE TRADES 55
4. A foundation wall for a building 28' 0" X 40' 0" is to be
7' 0" high and 1' 6" thick. There are to be 4 openings, two
3' 0" X 2' 6" and two 3' 0" X 5' 0". Concrete is to be used
in the construction and is to be mixed in the following pro
portions: — 1 cu. ft. (1 bag) of cement, 2\ eu. ft. sand and 5
cu. ft. broken stone. If \\ cu. yd. of dry material will make
1 cu. yd. of concrete, find the number of cu. ft. of cement, of
sand, and of broken stone.
5. A building 24' 6" wide, 36' 0" long and 20' 0" high, above
the foundation, is to be of stone with walls 16" thick. The
foundation, 6' 0" high, 16" thick, is to be concrete and to have
6 windows 1' 10" X 3' 4". If a cu. yd. of concrete requires 25
cu. ft. of stone, 12 cu. ft. of sand, and 4 cu. ft. of cement, find
the number of cu. ft. of each in the foundation. In the walls
of the house there are to be 8 windows 2' 0" X 5' 0", 3 windows
3' 6" X 5' 0" and 3 doors 3' 6" X 7' 0". How many cu. ft. of
stone will be necessary?
56. Brick Work. There is the same lack of uniformity in
methods of estimating cost in brick work as in stone work. In
measuring up the cost of the work some contractors make no
deduction for openings less than 2 ft. square. Usually, how
ever, the exact volume of the brick work is estimated and, in
fixing the cost, allowance is made for extra labour and material
for arches, cuttings, etc.
Since bricks are of varying size no fixed rule for the volume
of laid brick can be given. If we consider an ordinary stock
brick as 8f " X 2" X 4" and add a " joint to thickness, length
and width we get 9" X 2" X 4" or approximately 9" X 3"
X 4". The number of bricks for 1 cu. ft. of masonry would
then be 9^fiJ " 14 «
Table— (Based on above calculation).
Per cubic foot, 15 bricks.
Superficial foot of 9" wall, 11 bricks.
Superficial foot of 13" wall, 16£ bricks.
Superficial foot of 18" wall, 22 bricks.
The labour and material for brick work are usually estimated
by the 1000 brick, if in a straight wall.
56
MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises XXXV.
1. Make drawings to scale, in your laboratory book, of
bricks of different sizes. Allowing a §" joint calculate the
number of bricks that will be required for a wall 20' 0" long,
8' 0" high, and 18" thick.
2. A house is to have 27' 0" frontage, 30' 0" in depth, and
20' 0" in height above foundation. It is to have 8 windows
4' 6" X 5' 6" and 4 doors 4' 3" X 7' 0". The wall is to be 9"
thick; allowing 15 bricks to the cu. ft., how many bricks will
be required?
3. If the wall in the preceding question is 13" thick, find the
number of bricks.
4. What will it cost to lay the brick in each of the two
preceding questions if a bricklayer lays an average of 700 a
day and received 90c. an hour for an eight hour day?
5. The walls of a building 40' 0" wide and 100' 0" long are
to be 18' 0" high. There are 4 doors 8' 0" X 8' 0", 4 doors 3'
3" X 7' 0", 30 windows 4' 0" X 5' 0". Making use of the table
for superficial area find the number of bricks required, if the
wall is 13" thick?
6. Reckoning 15 bricks per cu. ft., find the cost at $30 a
thousand for the walls of a building 30' 0" wide, 50' 0" long
and 24' 0" high with the following specifications: — the lower
14' 0" is to have a wall 18" thick and is to have 4 doors 2'
10" X 6' 10" and 5 windows 3' 0" X 7' 0"; the upper 10' 0"
is to have a wall 13" thick, and is to have 6 windows 3' 0" X
5' 0".
57. Lumber. The common unit of measurement in lumber
is the board foot.
It is a piece of lumber
1 ft. long, 1 ft. wide, and
1 in. thick.
If we take a board
12 ft. long, 12 in. wide,
FlG  n and 1 in. thick, we
readily see that it will contain 12 board feet.
APPLICATION OF MEASURES TO THE TRADES 57
This might have been obtained as follows: — length (in
feet) X width (in feet) X thickness (in inches), thus 12 X 1
X 1 = 12.
This rule is applicable for finding the board feet of all kinds
of lumber. Example : — Find the number of board feet of lum
ber in a floor joist 2" X 10", 18' 0" long.
Solution: — Number of board feet = length (in ft.) X width
(in ft.) X thickness (in in.) = 18X^X2 = 30.
Lumber is billed in different ways,' (1) per thousand (M)
board feet, (2) per thousand (M) sq. ft., (3) per foot run.
Speaking generally we may say that, in dealing with
material 1" thick and up, the board foot is the unit, although
special sizes up to 2" X 3" are frequently charged as per foot
run. Below 1" in thickness material is reckoned in sq. ft.,
except "trim" which is sold as per foot run.
The following data for estimating the S S [3
amount of allowance for dressing and Fiq
working the tongue in flooring is furnished
by one of the large lumber companies of Toronto:
1§" wide, f " thick, add 50%
\\" wide, I* thick, add 33%
2" wide, I" thick, add Zl\%
2" wide, " thick, add 25%
2\" wide, " thick, add 33%
Example: — Find the cost of flooring a room 20' X 10'
with No. 1 red oak flooring, \\" X f ", at $160 per M
sq. ft.
Solution: — Area of floor = 20 X 10 sq. ft.
Lumber required = itnrX 20 X 10 sq. ft.
Cost = i£S X 20 X 10 X tVW = $48.00.
58 MATHEMATICS FOR TECHNICAL SCHOOLS
The following is a sample bill of lumber:
125
400
130
130
310
500
2000
2000
14
70
100
100
5
ft. lineal, lf"X6", Pine D4S.
ft. lineal, 1" XI", Pine Rgh.
ft. lineal, f" X10", Pine D4S.
ft. lineal, f " X2£", Bed Mldg
ft. lineal, f * Xlf", Fir Picture
Mldg
ft. B.M., 1" No. 1 H. D1S....
ft. Strip 6" H. Decking
ft. Strip  " Spruce Fig
pieces, 2"X 4' , X12 / , H. Szd.. .
pieces, 2" X 10" X 10', No. 1 H.
Szd
ft. lineal, 2" X 2", H. Rgh
ft. lineal, 3" X2", H. Rgh
pieces, 2"X5f"X10 / , Oak Sill.
125
400
130
130
310
500
2000
2000
112
1167
100
50
50
Price
1
8,
3.
3
62.
66
68
63.
65.
2.
65.
25
00
00
00
00
00
00
00
00
00
00
00
75
Amount
$ 9.06
4.00
10.40
3.90
9.30
31.00
132.00
136.00
7.06
75.86
2.00
3.25
37.50
Total.
461.33
D4S — dressed on four sides.
Rgh. — rough.
Mldg. — moulding.
Fig. — flooring.
Szd. — sized.
No. 1— No. 1 (best quality).
H. — hemlock.
Lin. — per foot run.
Exercises XXXVI.
1. Take measurements of a number of pieces of lumber
obtained from the woodworking shop. Make drawings in
your laboratory book and estimate the board feet in each.
2. Measure the top of a laboratory table, the top of the
teacher's desk, etc. Make drawings in your laboratory book
and estimate the board feet in each.
3. Take measurements of the floor of your classroom and
make a drawing in your laboratory book. Find the cost of
flooring with birch 2\" wide and §" thick at $140 per thousand
square feet.
APPLICATION OF MEASURES TO THE TRADES
59
4. By means of a drawing in your laboratory book, show the
number of board feet in a cubic foot.
5. Find the number of cu. ft. in a stick of timber
6" X 8" X 18' 0". Change to board feet and check by
rule.
6. An oak stick is 8" X 8" X 30' 0". Find its volume
by cubic measure. Change to board feet and check by
rule.
7. A lot 60' 0" frontage and 120' 0" in depth is to be enclosed
on two sides and an end by a tight board fence 6' 0" high.
The posts are to be placed 6' 0" apart and to cost 40c. each;
there are to be two string pieces 2" X 4" from post to
post on which to nail the boards; the boards are to be
1" thick. If lumber is worth $56 per M, find the total cost
of same.
8. What will it cost at $52 per M to cover the floor of a
barn 32' 0" X 42' 0" with 2" square plank?
9. A room is 12' 0" wide and 16' 0" long. Find the cost, at
$175 per thousand square feet, of laying a No. 1 red oak floor,
the material being \%" wide and f " thick.
10. Complete the following bill of lumber:
Feet
Price
Amount
500 ft. lineal, f "X3f", Pulley stile
500 ft. lineal, i"X3f ", Lining
$4.00
2.50
2.50
1.00
9.50
7.25
2.00
9.50
500 ft. lineal, 7"X3l", Lining
500 ft. lineal, Parting stop
150 ft. lineal, 2"X6", Sash sill
125 ft. lineal, lf"X6", Pine D4S
400 ft. lineal, \" X6", Backing
200 ft. lineal, 2" X6", Door jamb
Total.
Note — Prices are for 100 ft. lineal.
60 MATHEMATICS FOR TECHNICAL SCHOOLS
11. Complete the following bill of lumber:
Feet
Price
44 pieces, 2"X12"
10 pieces, 8" X 14"
105 pieces, 2" X 4"
15 pieces, 2" X 4 r 
32 pieces, 2" X 4"
17 pieces, 2" X 6"
23 pieces, 2" X 6"
9 pieces, 2" X 8"
20' 0'
16' 0'
long,
long,
10' 0" long,
12' 0" long,
8' 0" long,
16' 0" long,
12' 0'
14' 0'
long,
long,
Red Pine.
Red Pine.
Hem. Szd
Hem. Szd
Hem. Szd
Com. Pine
Hem. Szd.
Hem. Szd.
S84.00
86.00
63.00
63.00
62.00
86.00
65.00
66.00
Total.
12. Complete the following bill of lumber:
Price
12 pieces, 1"X7"16' 0" long, Pine D4S .
2 pieces, 6"X6"16' 0" long, Pine D4S..
310 ft. lineal, 8", Fir base
680ft. lineal, f "X2f ", Fir base D4S
46 pieces,  " X 5"14' 0" long, Door jamb
sanded
xlOOft. lineal, If "X3f ", Pine D4S
x9 pieces, If "X5f "10' 0"long, Pine D4S.
5 pieces, 2f " X5f "12' 0" long, Oak sill . .
5 pieces, 2f "X5f"10' 0"long, Oak sill. .
2 pieces, 2f "X5f" 14' 0" long, Oak sill. .
65 ft. lineal, 3", Crown moulding
xl20ft. lineal, l"X9f ", Clear PineD4S. . .
xl25 f t. lineal, f " X 5f ", Clear PineD4S. . .
xl25 ft. lineal, "X3f ", Clear PineD4S.. .
S 6.00
85.00
8.75
5.50
8.00
123.00
163.00
.75
.75
.75
3.75
190.00
160.00
150.00
Total.
x Dressed out of material even inch above.
APPLICATION OF MEASURES TO THE TRADES 61
58. Roofs, Rafters, Pitch.
In the above section of an ordinary gable roof, some of the
terms used in connection with roofs are indicated. The span
of a roof is the same as the width of the building. The run is
onehalf the span, and the rise is the vertical distance from the
top of the plate to the top of the ridge. The pitch of a rafter
is given by dividing the number of feet in the rise by the
number of feet in the span. Thus if the rise is 6 ft. and the
span 12 ft. the roof would have a onehalf pitch. The rafter
length is the distance from the outside corner of the plate to
the centre of the ridge. The heel is the distance from the
outside corner of the plate to the end of the rafter. The length
of the heel would have to be added to the rafter length if the
above method were used for the construction of the eaves.
62
MATHEMATICS FOR TECHNICAL SCHOOLS
The accompanying figures illustrate the method of finding the
lengths of the different rafters in a Hip or Cottage roof. Figure
Fig. 15
15 shows a plan of the roof, Figure 16 a right side elevation,
Figure 17 a plane at plate level.
In order to find the length of a hip rafter it would first
be necessary to find the length of HK in Figure 17. Using
APPLICATION OF MEASURES TO THE TRADES
63
this length and the perpendicular distance from H to the
ridge the length of the hip rafter may be found as in Figure 18.
Fig. 17
To find the length of the jack rafter we observe in Figure
P7 that, if the rafters be 16" on centre, MN would also be 16 ".
Fig. 18
Also since the roof has a \ pitch, the perpendicular distance
from the hip to N would also be 16", hence Figure 19.
If the roof has other than a \ pitch, similar triangles would
give the lengths of the jack rafters.
64
MATHEMATICS FOR TECHNICAL SCHOOLS
59. Roofing — Shingles. Shingles for roofing are estimated
as being 16" long and averaging 4" wide. They are put
up in bundles of 250 each, four bundles making a square of
shingles.
The unit in measuring for roofing is the square. A square
contains 100 sq. ft. If shingles are laid 4" to the weather,
each shingle would on an average cover an area of 16 sq. in.
This would give for 100 sq. ft. i^p or 900 shingles. In this
result, however, no allowance has been made for waste in
cutting or for defective shingles.
The following table has been found useful in practice
(Kidder's Pocket Book):
Inches to the
Area Covered by
Number to Cover
Weather
1000 Shingles
a Square
4
100 sq. ft.
1000
4J
110 sq. ft.
910
4
120 sq. ft.
833
5
133 sq. ft.
752
5£
145 sq. ft.
690
6
156 sq. ft.
637
60. Roofing — Slate. Slate for roofing is also measured by
the square (100 sq. ft.). In estimating either the amount
required or the cost of laying, eaves, hips, valleys, etc., are
measured extra — 1 ft. wide by the whole length. The sizes
of slates range from 9" X 7" to 24" X 14". "Each slate
should lap the slate in the second row below, 3 inches",
Kidder.
The gauge of a slate is the portion exposed to the weather,
which should be onehalf of the remainder obtained by sub
tracting 3 in. from the length of the slate.
APPLICATION OF MEASURES TO THE TRADES 65
The following table is taken from Kidder's Pocket Book:
Size of Slates
Inches Exposed
Number to a
in Inches
to Weather
Square
14X24
10i
98
12X24
10§
115
12X22
9
126
11X22
9*
138
12X20
8
142
10X20
8
170
12X18
n
160
10X18
71
' 2
192
9X18
n
214
12X16
6
185
10X16
6
222
9X16
6^
247
8X16
6
277
10X14
5*
262
8X14
5
328
Exercises XXXVII.
Note. — In working the following problems take the actual
quantity of lumber used, not allowing for waste due to having
to buy stock lengths of material. In case of fractional inches
take the inch above in each separate piece.
Lumber is cut in lengths of 10' 0", 12' 0", 14' 0", 16' 0",
18' 0", and will be charged on that basis.
1. Find the number of shingles for a square of roof for
each line in the table if no allowance be made for waste.
2. A shed 9' 0" wide and 18' 0" long is to have a "lean to"
roof, \ pitch. If the rafters are 2" X 4" at 16" centre and have
a 12" heel, find their cost at $52 per M. If the roof extends
12" on each end, find the cost of covering with 1" square
sheeting at $56 per M. Find the cost of shingling the above
with shingles laid 4" to the weather, if material and labour
cost $14 a square of shingles.
3. A garage 10' 0" wide and 16' 0" long is to have a gable
roof, \ pitch. The rafters are 2" X 4" at 2' 0" centre and have
a 15" heel. The rafter ties are 2" X 4" X 10' 0". Find the
cost at $50 per M. If the roof extends 10" on the ends and
66 MATHEMATICS FOR TECHNICAL SCHOOLS
6" more on each end be allowed for waste, find the cost of
covering with 1" square sheeting, at $48 per M.
Find the cost of shingling the above with shingles laid
4?" to the weather if material and labour cost S13..50 per square
of shingles.
4. A stable 15' 0" wide and 20' 0" long is to have a gable
roof, \ pitch. The rafters are 2" X 4" at 20" centre and have
an 18" heel, the ridge board
being 1" X 6". The roof is
supported at every second
rafter by a brace 2" X 4"
(see Figure 20) and collar
ties 2" X 6" X 15' 0". Find
the cost of lumber at $52
per M. If the extension on
the ends is 12", find the cost
of sheeting with 6" tongued
and grooved lumber at $55 per M sq. ft., allowing 10% for
the tongue and groove and 6" on each end for waste.
Find the cost of shingling the above roof with shingles laid
5" to the weather if material and labour cost $13 per square
of shingles.
5. A house 25' 0" wide and 32' 0" long is to have a gable
roof, f pitch. The rafters are 2" X 6", 16" on centre, with an
18" heel. The roof is supported by braces 2" X 4", 4' 0" on
centre, 6' 0" long, and tied with ceiling rafters 2" X 6"X25'0".
Find the cost of the above lumber at $55 per M.
If the extension on the ends be 12", find the cost of covering
with " X 6" tongued and grooved sheeting at $66 per M sq.
ft., allowing 10% for the tongue and groove and 8" on each end
for waste.
Find the cost of roofing the above with slate, the gauge
being 8^", if material and labour cost $30 a square.
6. A building 20' 0" wide and 28' 0" long is to have a hip
roof, \ pitch, the ridge being 1" X 8" X 8' 0" long. The
hip rafters are 2" X 6", the jack rafters 2" X 6", at 16"
centre. Ceiling joist 5' 0" from plate level, 2" X 6"; act as
ties. Find the cost of the above lumber at $56 per M.
Note. — In estimating the amount of material in rafters, find
the length of a common rafter and multiply by the number
of rafters on both sides.
APPLICATION OF MEASURES TO THE TRADES 67
7. A building 22' 0" wide and 32' 0" long is to have a hip
roof,  pitch, the ridge being 1" X 8" X 10' 0". The hip
rafters are 2" X 6", the jack rafters 2" X 6", at 16" centre;
ceiling joist 2" X 6" X 22' 0" act as ties. There are also 14
braces 2" X 4", at 16" centre, from centre of ceiling joist to
centre of common rafters. Find the cost of lumber at $55
per M.
61. Lathing and Plastering. In lathing and plastering the
square yard is the unit of measurement.
Standard laths are 4 ft. long, 1\ in. wide, and are laid \ in.
apart. They are usually put up in bundles of 50. It requires
approximately 18 laths to cover a square yard. In estimating
both the amount and the cost of lathing and plastering, the
percentage of the openings deducted from the total area will
depend upon the job. In small openings no deduction will be
made; in medium openings about 40% or 50%; in very large
openings from 75% to 90%.
Exercises XXXVIII.
1. What will it cost to lath and plaster the walls and ceiling
of the following rooms at 70c. per square yard? (a) 17' 0"
X 13' 0" X 9' 0" high, with a door 3' 0" X 7' 0", and 3 windows
each 3' 3" X 5' 0". Deduct 40% of the openings. (b) 20'
0" X 18' 0" X 11' 0" high, with 2 doors 3' 0" X 9' 0", and 4
windows 3' 0" X 5' 0". Deduct 50% of the openings, (c)
40' 0" X 22' 0" X 15' 0" high, with 2 doors 5' 6" X 8'0", and
window space 15' 0" X 9' 0". Deduct 90% of the openings.
2. A room is 16' 0" wide, 18' 0" long, and 10' 0" high. There
are 2 doors 4' 6" X 7' 0", 3 windows 3' 8" X 4' 9", the sills
being 2' 3" from the floor. Find (a) the cost of lathing
and plastering at 90c. a sq. yd., deducting 50% of the openings;
(b) the cost of laying and finishing a " X 2" No. 1 red oak
floor at 32c. per sq. ft.; (c) the cost of paneling the walls to a
height of 4' 0" at 80c. a sq. ft. (Note — plastering is carried to
floor behind paneling.)
3. Find the cost of lathing and plastering a room 27' 0"
wide, 30' 0" long, and 12' 0" high at 80c. a sq. yd., if there are
2 doors 3' 6" wide and 7' 0" high, and 6 windows 3' 4" wide and
6' 0" high. Deduct 50% of the openings and also deduct 12c.
a sq. yd. for the area paneled (see No. 4), on account of the
finishing coat being unnecessary.
68 MATHEMATICS FOR TECHNICAL SCHOOLS
4. Find the cost of paneling the walls in the preceding
question to a height of 4' 6" if the sills of the windows be 2'
6" from the floor, at 85c. a sq. ft.
5. A hall is 50' 0" wide, 90' 0" long, and 20' 0* high. There
are 4 doors 5' 6" X 10' 0", 2 windows 5' 0" X 11' 0", 7 windows
5' 0" X 8' 3". Find the cost of lathing and plastering at 75c.
a sq. yd., deducting 80% of the openings.
62. Decorating and Painting. Wall paper is put up in
rolls, the number of yards in the roll and the width of the
paper varying. The kinds chiefly in use are:
(1) Paper 18" wide and in single rolls 8 yd. in length, or
double rolls 16 yd. in length.
(2) Paper 21" wide and in rolls 12 yd. in length.
(3) Paper 30" wide and in rolls 5 yd. in length; frequently
put up in 15 yd. rolls.
To estimate for the walls.
(1) Find the perimeter of the room, less the width of doors
and windows.
(2) Find the number of strips required by dividing the result
in (1) by the width of the paper.
(3) Find the number of strips that can be cut from a roll by
dividing the length of the roll by the height to be papered.
(4) Find the number of rolls by dividing the number of
strips required for the room by the number of strips in a roll.
To estimate for the ceiling.
If the strips are to run lengthwise, find the number of strips
by dividing the width of the room by the width of the paper,
then proceed as in case of walls.
To estimate for the border.
Find the total perimeter of the room. Estimate cost per
running yard.
When double rolls are available they would be used, if
more economical in cutting.
APPLICATION OF MEASURES TO THE TRADES 69
Example:— A room 16' 0" wide, 20' 0" long and 9' 0" high
from baseboard, has two doors each 4' 0" wide and three
windows each 3' 6" wide. Find the cost of paper for the walls
and ceiling, the wall paper being 18" wide and costing $2 a
double roll, the ceiling paper being 18" wide and costing 80c.
a double roll.
Perimeter of room = (16' + 20') 2 = 72'.
Perimeter — Width of doors and windows = 72' — 18£' = 53?'.
Number of strips required = — ^^ — = 35f .".36.
Number of strips in a double roll= ^ = 5^ .'. 5.
Numberof rolls *£ = 7£.\8. Cost = $16.
Number of strips required for the ceiling if running length
16X12 in2 .
wise = — r^ — = lOf . . 11.
lo
Number of strips in a double roll = f£ = 2f .'.2.
Number of rolls = V = o§ .'.6. Cost = $4.80.
Total Cost $20.80.
Painting. The area is usually estimated in sq. yd.
The following is a common method of reckoning the area
of doors, windows, etc.:
Doors are taken to average 3' 0" X V 0", windows 3' 0"
X 6' 0". If the window be divided into 12 lights the area is
doubled, if divided into 6 lights onehalf the area is added,
and so on. The baseboard is taken as 1' 0" by total perimeter,
picture moulding 3" by total perimeter, and dado rail 6" by
total perimeter.
Exercises XXXIX.
1. A room is 13' 0" wide, 15' 0" long and 8' 6" high. There
are two doors each 2' 8" wide, and three windows each 3' 0"
wide. Two of the windows have 6 lights and the other 2
lights. A picture moulding 2" wide and a baseboard 8"
wide run around the room. Find (1) the cost of tinting the
ceiling and 1' down to picture moulding at 25c. a sq.yd., (2)
the cost of painting the interior woodwork at 50c. a sq. yd.,
70 MATHEMATICS FOR TECHNICAL SCHOOLS
(3) the cost of papering the walls with paper 21" wide at
$1.25 a roll, the decorator charging 40c. a roll for the work.
2. A room 10' 8* wide, 11' 4" long and 8' 6" high, has two
doors each 2' 10" wide, one window 4' wide, 2 lights, two
windows each 3' wide, 12 lights, baseboard 10" wide running
around the room. Find (1) the cost of painting the woodwork
at 25c. a sq. yd., (2) the cost of papering the ceiling with
paper 18" wide at 25c. a single roll, (3) the cost of papering
the walls with paper 30" wide at 90c. a roll, using a border
4" wide at 20c. a yard. The decorator charges 30c. a roll for
the work in both walls and ceiling.
3. A room 12' 0"'wide, 18' 6" long and 10' 0" high, has two
doors each 3' 10" wide, two windows each 2' wide, 4 lights in
each, one window 4' 6" wide, 12 lights, a fireplace 5' 6" wide,
a picture moulding 3" wide and a baseboard 1' 0" wide running
around the room. Find (1) the cost of tinting the ceiling and
16" on wall to picture moulding at 30c. a sq. yd., (2) the cost
of painting the woodwork at 40c. a sq. yd., (3) the cost of
papering the walls with paper 18" wide at $1.20 a double roll,
the decorator charging 50c. a roll for the work.
CHAPTER VI.
ALGEBRAIC NOTATION.
63. In Arithmetic we denote quantities by numbers, each
number having a fixed value. By 5 in. we mean that the line,
or pencil, or bolt, is 5 in. in length. For this purpose we have
the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These symbols, in
whatever way they are combined, have definite fixed values.
In Algebra there is no limit to the number of symbols employed,
the letters from our own alphabet being the ones chiefly
used.
64. Algebra — generalized Arithmetic. These symbols: — a,
b, c, d, etc., in contrast to the symbols of Arithmetic, have not
fixed values but may be given any values required by the
conditions under discussion. Thus in Arithmetic 2X4 is
always 8, whereas 2Xa, or more briefly 2a, will have different
values according to the numerical values assigned to the
symbol a. When a = 4, 2a = 2X4 = 8, when a = 8, 2a = 2X8
= 16, and so on. Here the 2a is called an Algebraic Expression,
the 2 being called the Numerical Coefficient, denoting the
number of times a is taken in the sum. When no numerical
coefficient is placed in front of the symbol, 1 is understood, thus
a means la.
65. Arithmetic Laws Applicable. Since the symbols a, b,
c....x, y, z stand for numerical quantities we may apply
the ordinary Arithmetic laws infusing them. In Arithmetic
2X6+3X6 = 5X6 = 30. So in Algebra 2a+3a = 5a, 6a2a =
4a. In Arithmetic 2X6 = 6X2, so in Algebra aXb = bXa.
Also 2X4X6=4X6X2 = 6X2X4, so in Algebra aXbXc =
aXcXb = bXaXc = abc = acb = bca. If we then wish to add
3afec+2ac6+7ca6 we should rearrange the terms thus,
3abc+2abc+7abc = 12abc. An important difference between
n
72
MATHEMATICS FOR TECHNICAL SCHOOLS
the notation of Arithmetic and that of Algebra should be
noted. In Arithmetic 31 means thirtyfour or 3X10+4; in
Algebra ab means aXb.
Exercises XL.
Find the values of the following:
1. 3x+9x.
2. a\a.
3. 2a — a.
4. 7x3x.
5. llx— 4a\
6. x— x.
7. 3a6+5a6.
8. 2a6+36a.
9. ab — ba.
10. Wxy — lxy.
11. 9xy — 3yx.
12. 6a6 — 6a.
13. 8a6c — Scab.
14. 3x+4x+5x.
What is the value of 8x when:
21. x = 2.
22. x = 4.
23. x= i
24. x=4.
What is the value of ^ when :
27. z= 4.
28. a; = 16.
29. x = 5.
30. x = \.
What is the value of ^ when:
o
33. x= 6.
34. x = 18.
35. x = 75.
36.' a: = 27.
15. 3a6+2a6+4a6.
16. 5xy +Qxy+3xy.
17. 3a6c+26ca+10ca6.
18. x\x+x+x.
19. 3x+4x+x+6x.
20. 96+36+56+66.
25. x= I
26. x = 2\.
31. x= 5.
32. x = 2b.
37. ar=6.
38. x=9.
39. x= 036.
40. x=0024.
Exercises XLI.
1. What is the number which is 2 greater than x?
2. What is the number which is 3 less than x?
3. If an article costs x cents what is the cost of three articles?
of seven articles? of eleven articles?
4. Express x sq. ft. in sq. in.
5. Express x sq. in. in sq. ft.
6. Express x metres in (1) decimetres, (2) in centimetres,
(3) in millimetres, (4) in kilometres.
7. Express x millimetres (1) in centimetres, (2) in decimetres,
(3) in metres, (4) in kilometres.
ALGEBRAIC NOTATION
73
8. If there is an average of x trains leaving Toronto every
day and an average of y cars per train, how many cars leave
Toronto per day?
9. In a rectangle ABCD if AB is c ft. in length and BC, b ft.
in length, find (1) the perimeter of the rectangle, (2) the area
of the rectangle.
10. If the side of a square is b feet, find its perimeter.
11. A can do a piece of work in m days and B in n days;
write down (1) the amount of work each can do in 1 day, (2)
the amount of work both can do in 1 day.
12. The sides of a triangle measure x, y, z ft. Write
down an expression for (1) the perimeter, (2) the semi
perimeter.
13. A merchant mixes x ib. tea worth z c. a lb. with
n lb. worth yea lb. Find the value of one lb. of the
mixture.
14. If a man works x hr. per day and handles y castings
per hour, how many castings does he handle each day?
15. If there are x cars in a railroad yard, how many
trains will there be if there is to be an average of b cars
per train?
16. What is the length of the casting in the accompanying
figure?
Fig. 21
17. What is the length of the casting in the accompanying
figure?
<f—
Fig. 22
74
MATHEMATICS FOR TECHNICAL SCHOOLS
18. Find the length of the crankpin in the accompanying
figure.
<
s«
«.
r
 W. J
w »
r
■
*
J
Fig. 23
19. If I is the length of the crankpin in the accompanying
figure what is the length of the last step?
I
c
s
^
f
r
<'
J
Fig. 24
20. If I is the length of the cylinder and saddle shoulder
bolt in the accompanying figure what is the length of the
shoulder?
Fig. 25
66. Index or Exponent, Power. In Arithmetic 3X3 may
be written 3 2 or 9.
In Algebra if we multiply a by a we cannot write the product
as a single symbol, since we do not know the value of a; but
we may express it as a 2 . In a similar way aXaXa^a 3 ,
aXaXaXa^a*. The small figure placed to the right and above
the symbol is called the Index or Exponent, and the product a*
is called the fourth power of a or more commonly a to the
fourth.
ALGEBRAIC NOTATION 75
67. Index Laws. x i Xx 5 = xxxxxxxXxxxxxxxxx = x 9 .
From this example we have the law: — The index of the
product of two powers of the same symbol is equal to the sum
of the indices of the factors.
Examples : — a 6 X a 5 = a £+ 5 = a 11 . x 3 Xx 7 = x 10 .
a 2 bXb 2 a = a 2+1 Xb 1 + 2 =a 3 b 3 .
, . a 5 aXaXaXaXa
a 5 + a 3 = —. = — — = a 2 .
a 6 aXaXa
If we cancel 3 of the factors in the numerator by the 3
factors of the denominator, the above expression becomes
aXa = a 2 . From this example we infer the law: — The index
of the quotient of one power of a symbol divided by another power
of the symbol is obtained by subtracting the index of the divisor
from the index of the dividend. Examples:
a 7 + a* p a 1 ~ i = a 3 . 6 25 r6 14 = 6 25  14 = 6 u .
68. Some expressions in detail, a 3 means that a is taken
three times as a product.
ab 2 means that a is taken once, b is taken twice as a product,
and the two results are multiplied together.
6a 2 6 3 means that a is taken twice as a product, b is taken
three times as a product, the two results are multiplied together,
and the resultant product is taken six times.
Exercises XLII.
What is the:
1. second power of a? 6. product of a 2 and a 3 ?
2. third power of 4? 7. product of 4a and 36?
3. fifth power of 5? 8. product of 4a 2 and 5a 3 ?
4. sixth power of 6? 9. product of 12a6c and 3a6c?
5. product of x and x 2 1 10. product of 3a 2 6 and a& 2 ?
11. Express the product abx 2 in different forms.
12. Do the same with Sx 2 y 3 , 6a 2 6 3 c 4 , 12ab 3 x.
13. Write in detail what is meant by the following:
4a 3 6, 5a 2 6 2 , 6a6c, 7a 3 6c.
76 MATHEMATICS FOR TECHNICAL SCHOOLS
Find the results of the following expressions in the most
simplified form :
14. a+3af6a. 16. 3a6c+6a6c.
15. 7a2a+4a. 17. 7a6 + 106a.
18. 13a6c+66ca — 2cab.
19. 4:xyz{Qyxzt2zxy — £zyx.
What is the result of:
20.
a 6 4 a 2 .
24.
xyrx^y 4 . 28.
a 2 b 2 + abXa 3 b 3 .
21.
x 16 ix 3 .
25.
3a 2 6 2 Ha6. 29.
Sx 3 y 3 7xyX 2x 2 y 2 .
22.
x 3 y 3 ±xy.
26.
15x 3 i/ 3 7a;V. 30.
4:m 3 n 3 X rnn + wi 2 n 2 .
23.
x 2 +x 6 .
27.
x 2 y 2 Xx 3 y 3 sxy.
31. The side of a square is 6 in. What is its area?
32. The edge of a cube is b in. What is the area of a face?
What is the area of all the faces? What is the volume of the
cube?
33. The volume of a cube is 8x 3 . What is the area of a
face? What is the area of all the faces?
34. If a train travels I hr. at k miles per hr. and c hr. at d
miles per hr., find the total distance travelled.
35. Represent three consecutive numbers, (1) if x is the
first one, (2) if x is the middle one, (3) if x is the last one.
36. If the length of a stick is b ft. find its length in in.,
in yd., in rods.
37. If a rod is x yd. b ft. and c in., how many inches in length
is it?
38. If x is the price per quart for beans, what is the price
per gallon? What is the price per bushel?
39. A man earned $x per day and his son $y. How many
dollars did they both earn in a month if the man worked 25
days and the son 20 days?
69. Roots. As in Arithmetic the square root of x, or the
expression whose second power is x, is indicated by y/x.
Similarly the cube, fourth, fifth, etc., roots of x, or the expres
sions whose third, fourth, fifth, etc., power is x, are indicated
by s/x, \Zx, \Zx, etc.
ALGEBRAIC NOTATION
77
Thus, ^a 6 =a 2 Since a 2 X a 2 X a 2 = a 6 .
^a 12 = a 3 Since a 3 X a 3 X a 3 X a 3 = a 12 .
^32=2 Since 2X2X2X2X2 = 32.
The symbol V is called the radical sign.
Exercises XLIII.
Find the square root of:
1. x 2 . 6. 16a 6 .
2. x 6 . 7. 49x 2 # 2 .
3. 16x 2 . 8. 81a 4 6 4 .
4. x 12 . 9. 144a 6 6 6
5. 64x 4 . 10. 169aV
Find the value of:
16. y/a 4 b\
17. V« 6 .
18. V49V36.
19. V49 + V4.
20. Vx 6 x 3 .
21. y/x*x\
22. ^x s .
23. ^/x 16 .
11.
12.
13. ~
a?
4*
a 6
9'
x 8
f6'
14.
15.
a 2 b 2
9 "
x 4 y 4 z 4
16 '
25. Square of 4x?/.
26. Cube of x 2 .
27. Fourth power of y 2
28. Cube of 2 a 2 ?/ 4 .
29. Cube root of x 6 .
30. Cube root of 8a 3 .
31. C ube root of 27a 6 .
32. V2516.
34. V25a 4 16a 4 .
35. Vfi
36 €
24. Square of a 4 6. 33. V4933.
\a 16 .
38
70. Like and Unlike Terms. Two terms which contain the
same letters involved in the same way are called like terms.
Thus 6a and 3a are like terms. 3a6 and 4a6 are like terms.
7x 2 and 9x 2 are like terms.
Since ab and ba both mean aXb, ab and ba are also like
terms, also 5a6 and 76a are like terms.
Like terms may then be defined as terms that differ only in
their numerical coefficients.
Unlike terms may be defined as terms that differ in other than
their numerical coefficients.
Thus 6a and 46 are unlike terms. x 2 y and xy 2 are unlike
terms. 7a 2 and 96 2 are unlike terms.
78 MATHEMATICS FOR TECHNICAL SCHOOLS
If we wish to add such terms all we can do is to write them
down with a plus sign between them, thus 6a +46, x 2 y\xy 2 ,
7a 2 +96 2 .
When we wish to simplify an algebraic expression such as
3a+46 — 2a +66 we can combine the like terms 3a and — 2a,
giving a, and the like terms 46 and 66, giving 106, and write
the result a+106.
Examples :
10a+663a+4c26c = 10a3a+6626+4cc
= 7a+46+3c.
9xy + 4x 2 ?/ 2 + 2xy — 3x 2 y 2 = 9xy + 2xy + 4x 2 2/ 2 — Zx 2 y 2
= llxy\x 2 y 2 .
Exercises XLIV.
Simplify by combining like terms:
1. 4a+3a+6a2a. 5. 7xy+6x 2 y 2 3yx+4y 2 x 2 +3xy 2x 2 y 2 .
2. 3a+2a + 66 — 46. 6. 3ra + 2rc+2m — m — n\3mn — n\2mn.
3. 3a6+46a+36c6c. 7. 6p+2g+4r3p+6g2r+4p2g.
4. 6a6c+3a 2 6 2 26ca. 8. 3a+2x4?/+7a+82/+5x.
If a = 8, c = 0, k = 9, x = 4, y = l, find the value of:
VcyK 12. 2x^2ay.
25a
9.
V2ak 2 .
10.
yzk.
11.
13.
5y\/4:kx.
14.
Sc\/kx.
15.
17.
18.
Ikax 2
\i8y.
v^ v 2
71. Brackets. In Arithmetic when a number of terms is
included within a bracket it is understood that these terms
are to be regarded as a whole.
Thus, 10+ (5 +4) means that we first add 5 and 4 and then
add the result to 10. Also 10 — (5+4) means that we first
add 5 and 4 and then subtract the result from 10. So in
Algebra, a + (6+c) means that we first add 6 and c and
then add the result to a.
ALGEBRAIC NOTATION 79
Certain rules are necessary with respect to the signs of the
terms within the bracket when the bracket is removed. These
rules may be obtained by an analysis of a few type cases.
By a\(b{c) we mean that the quantity b\c is to be added
to a. We may first add 6 and then afterwards add c, giving
a\b+c. By a\(b — c) we mean that the quantity obtained
by subtracting c from b is to be added to a. It is evident
that if we add b to a, obtaining a\b, our result will be too
great by c; we must therefore subtract c from a\b, obtaining
a\b—c as a result. From these illustrations we infer the
rule: — When a group of terms is contained within a bracket
preceded by the sign + the bracket may be removed without
changing the signs of the terms within.
In a — (b\c) we have to subtract the sum of b and c from a.
If we subtract b from a, giving a — b, it is evident that the
result is too great and that it is too great by c; therefore we
must subtract c from a — b, giving a — b — c. In a — {b—c) we
have to subtract the result b — c from a. If we subtract b
from a, giving a — b, it is evident that we have taken away too
much, for we were required to take away only b — c. The
result a — b is therefore too small by c, and we must add c to
a — b, giving a — b\c. From these illustrations we infer the
rule: — When a group of terms is contained within a bracket
preceded by the sign — the bracket may be removed by changing
the signs of the terms within.
3x means x\x{x, similarly 3(af6) means (a+6) + (a+6)
+ (a+6) =3a+36. This would lead us to the rule: —
The product of an expression, consisting of two or more
terms and a single factor, is the sum of the products of each
term of the expression multiplied by the single factor.
Examples: 1. 3x — (a\b) =3x —a —b.
2. la +(b+c)=7a +b +c.
3. 9x 2 — (x— y) =9x 2 — x \y.
4. Q(a+b+c) =6a +6b+Qc.
80 MATHEMATICS FOR TECHNICAL SCHOOLS
It is necessary to note the difference between 3a 2 and
(3a) 2 . In 3a 2 we have to multiply a by a and take the result
three times. In (3a) 2 we have to square the whole quantity
3a, giving 3aX3a or 9a 2 .
Examples: 1. (7a6) 2 = 7a6X7a6=49a 2 6 2 ,
2. (2a 3 ) 4 =2a 3 X2a 3 X2a 3 X2a 3 = 16a 12 .
It is sometimes necessary to enclose with brackets part of an
expression already enclosed within brackets. In such cases
the pairs of brackets are made of different shapes — ( ),
{},[]. Thus a{6 + (cd)}.
The same rules with respect to the removal of brackets
apply, it being usually best to begin with the inside pair and
remove one pair at a time. In the example given we would
first simplify thus, a— {6fc— d}. We would then remove the
remaining pair and write the expression a — b — c+d.
Simplify:
Exercises XLV.

1. 3a + (4a
2. 15x(6a
2a).
:+3x).
3.
4.
36
6a
(2c
(4c
t+4a).
t + 2a).
Prove the following by removal of brackets:
5. 6r0r2)(3+4x) + (6x + l)=3.rf2.
6. (3x2)(4x+5) + Or+7)=0.
7. (9a b) + (3a 26) (6a 56)= 6a +26.
8. (xr6a)(2a:3a)(a6x)=5x(8a.
9. 2(x + l)+3(l+x)+2(2+3x)=9 + llx.
10. 3(2a)+6(2a+7) + (a42) = 10a+6.
11. 2(a+6)(2a6)=36.
12. 3(a+brc)(6+ac)(2c2a6)=4a+3&+2c.
13. 2(3x+12)f3(xf4)(8a;12)=x+48.
Simplify:
14. 3{x(2x6x)}. 17. 3x 2 +x(x+S)+x 2 .
15. {3a + (6a2a)+4a}. 18. a {6 + (c6?)}.
16. x+{2zf30r+2x)}. 19. a[6 {a(6a)f6} a]
ALGEBRAIC NOTATION 81
20. Enclose a — b+c — d — e+i in alphabetical order in
brackets, two letters in each; three letters in each.
72. Negative Quantities. We have in Arithmetic found
the value of an expression such as 6 — 5. In every case however
the number to be subtracted was less than the number from
which it was subtracted. ,
A difficulty is presented if we are asked to find the value of
5 — 6. This is arithmetically impossible. We cannot take
$6 from $5. We may, however, by making use of brackets,
write 5 — 6 thus, 5 — (5 + 1) =5 — 5 — 1 . Here 5 — 5 is 0, and the
value appears as 1 to be subtracted with nothing from which
to subtract it. We shall say that the result is the negative
number 1 or minus 1, and denote it by — 1. The idea of negative
numbers may be made clearer by means of a graphical repre
sentation.
I + + + + + +
654321 123456
In the above diagram we have represented numbers to the
right of the vertical line as positive, and numbers to the left
of the vertical line as negative. The two series of numbers
may be considered as forming but a single series consisting of a
positive branch, a negative branch, and zero.
If then we wish to subtract 4 from 2 we begin at 2 in the
positive series, count 4 units in the negative direction (to the
left) and arrive at — 2 in the negative series, that is, 2 — 4 = 2.
A few examples may be added to show the practical value
of negative quantities.
Example 1 : — If the temperature is 30° below zero it may be
recorded 30°. If it rises 5° it is then 25° below zero or 25°.
If it increases 10° more it is 15° below zero or — 15°.
Example 2: — If a merchant during a day's transactions gains
$80 on one class of goods and loses $100 on another class we
can represent the result of the day's business as $80 — $100 =
$20.
82 MATHEMATICS FOR TECHNICAL SCHOOLS
Example 3 : — If a man rowed 50 yards up stream and then
drifted down 60 yards, his position relative to the starting
point would be 50 yards — 60 yards = — 10 yards.
Exercises XL VI.
1. A man has $500 and owes $500. How much is he worth?
2. A man has $500 and owes $700. How much is he worth?
3. A man goes 5 miles north of Barrie, then 9 miles south.
How many miles north of Barrie is he? How many miles
has he travelled? Make a diagram showing his route and his
last position.
4. The temperature at 6.00 A.M. is +14° and during the
morning it grows colder at the rate of 4° an hour. Find the
temperature at 9.00 A.M., at 10.00 A.M., and at noon.
5. A freight engine is switching in front of a station. If
it runs 400 ft. to the right of the station (+400 ft.) and then
backs 525 ft. (— 525 ft.), how many feet is it from the
station?
6. In drilling a well the drill is raised 8 ft. (+8 ft.) above the
surface. It is then dropped 15 ft. ( — 15 ft.). Where is it
then with respect to the surface?
7. A boy is fishing in deep water with a line 20 ft. long. If
the tip of the pole is +6 ft. above the water, how far is the
sinker from the surface of the water, if it is 3 ft. from the hook?
8. A man who was $350 in debt contracted another debt
of $200. He then earned $1000. How much was he then
worth?
9. A boat, that runs 16 miles an hour in still water, is going
against a stream flowing 4 miles an hour. What is the rate
at which the boat travels?
10. If a mine is opened 200 ft. above the base of a mountain
and a shaft is sunk 700 ft., how much is the base of the shaft
above or below the base of the mountain?
11. A man starts from a point on a road running north
and south, and walks c miles north and then 6 miles in the
opposite direction. How far is he now from the starting
point? How far has he travelled?
Illustrate from the following cases: — (1) c = 8, 6 = 6.
(2) c = 5, 6 = 9. (3) c = 8, 6 = 8.
ALGEBRAIC NOTATION 83
12. The thermometer stands at x°; in the course of an
hour there is a fall of y° and in the course of the next hour a
rise of z°. Find the reading at the end of this time.
Illustrate for the following cases: — (1) x = 6, y — 4, z = 5;
(2) x = 8, = 4, z = 9; (3) x= 2, = 3, z = 6.
Simplify:
13. 3a+2&4a+668a9&.
14. 2s3s+ss5s+5s.
15. 4.r 3 ox 3 +3x 3 8x 3 +7x 3 .
16. 56 +§6f& +26 !&+£&.
17. fx0+a;+f0x+00.
18. 6a 2 3a 2 +2& 2 +3a 2 2& 2 .
19. x 2 +xy+y 2 3x 2 2xy+4y 2 .
20. 3p+252p+4g6p.
CHAPTER VII.
SIMPLE EQUATIONS.
73. We might say that the greater part of a student's work
in Arithmetic has been concerned with equations. The state
ment that 3 added to 4 is 7 might be expressed in the form
34 = 7. This is an equation or a statement of equality
between two expressions, 3+4 being one and 7 the other.
All such equations involving only simple Arithmetical
operations may be called Arithmetical equations, to distinguish
them from equations of the form Sx = 9 which we will call
Algebraic equations.
The x in this equation is called the unknown and the process
of finding its value is called solving the equation.
An equation, in which the unknown quantity is involved to
the first power only, is called a simple equation.
In the given case if 3x = 9, then x = 3; the value 3 is said to
satisfy the equation.
D"
O
Fie. as
74. Operations on the equation. The two sides of an
equation must always balance, just as the weights in the
two pans of the scales above must be equal if the scales are to
84
SIMPLE EQUATIONS 85
balance. In the equation 3x = 9 if we add 2 to the lefthand
side we must of necessity add 2 to the righthand side. The
equation then becomes 3x +2 = 9 +2 or 32+2 = 11.
In the same way, if we subtract 2 from the lefthand side
we must subtract the same quantity from the righthand
side. The equation then becomes 3a: — 2 = 9 — 2 or 3x — 2 = 7.
Further, if we multiply the lefthand side by 2 we must
multiply the righthand side by 2, giving 2X3z = 2X9 or
Qx = 18.
We might also divide the lefthand side by 2 giving fx,
but we would also have to divide the righthand side by 2,
giving f or the equation f#=f.
75. Transpositions. Let us consider the equation 3x+4 = 16.
By the previous paragraph we could subtract 4 from both
sides of the equation, giving 3x+4 — 4 = 16— 4 or 3x = 16— 4.
We then observe that the equation 3x+4 = 16 is equivalent
to 3x = 16 — 4, or that the 4 has been moved from the lefthand
side to the righthand and that its sign has been changed.
Next let us consider the equation Sx — 2 = 10. We could add
2 to both sides giving 3x2+2 = 10+2 or 3x = 10+2. We
then observe that the equation 3^ — 2 = 10 is equivalent to
3z = 1042, or that the 2 has been moved from the lefthand
side to the right and that its sign has been changed. These
two examples would lead us to make the statement: — A quantity
may be transferred from one side of an equation to the other
without altering the balance, provided we change the sign of the
quantity transferred.
Examples:
1. Solve the equation 3x2+5x4 = 3:r107x + 16.
Transposing so that we have all the terms containing x on
the left and the other terms on the right we get
3x+5x3x+7x= 10 + 16+2+4 giving 15x3x = 2210.
or, 12x = 12.
x = l.
86 MATHEMATICS FOR TECHNICAL SCHOOLS
2. Solve the equation 3(3z + l) (x1) = 6(x + 10).
Multiplying out 9x+3x + l =6x + 60.
Transposing 9x— x — Qx = 60 — 3 — 1.
2x = 56.
x = 28.
Verification: — Substitute the value 28 for x in the equation
and we get 3(84 + 1) (281) =6(28 + 10)
or, 3X8527 = 6X38
25527 = 6X38
228=228.
76. Need of the equation. Let us work the following
problem, first without employing Algebraic symbols and
then by making use of the Algebraic symbols, and compare
the methods.
Example :
A shopper bought three articles, the second costing three
times as much as the first and the third $3 more than the
second; find the cost of each if the total cost was $10.
First solution: — Suppose that the third article had cost as
much as the second, then the total cost would have been
$10— $3 or $7. Then for every share allotted to the first article
we must allot three to the second and three to the third. This
makes seven shares into which we must divide $7, giving $1
for one share.
/. the first article cost $1,
the second article cost $3,
the third article cost $3 +$3 = $6.
Second solution: — Let x = No. of dollars in cost of first,
then 3z = No. of dollars in cost of second,
and 3x+3 = No. of dollars in cost of third,
thena;+3a;+3a:+3 = 10
7z+3 = 10
7x = 103
7x = 7
x = l
3x = 3
3z+3 = 3+3 = 6.
SIMPLE EQUATIONS 87
or the first article cost $1, the second $3, the third $6. If we
compare the two solutions it is evident that the latter method
has the advantage in both directness and clearness.
Additional examples:
1. A man works a full day of 8 hours, and in addition works
3 hours overtime, for which he receives time and a half. If he
is paid $8.75 for the entire time, what is his regular rate per
hour?
Let x c. = regular rate per hour,
then f x c. =rate per hour for overtime,
then 8 x c. =pay for 8 hours' work,
and 3 Xf xorfx c. =pay for overtime,
.'. &r+fa: = 875.
Multiplying both sides of the equation by 2 we get
16x+9x = 1750
25x = 1750
a; = 70,
or the regular rate is 70c. per hour and the overtime rate is
 X 70 or $1.05 per hour.
2. How much water must be added to a quart of alcohol,
which already contains 5% of water, so that the mixture may
contain 50% of alcohol? (No allowance being made for con
traction).
Let x =the number of quarts of water to be added,
then 1+x = total number of quarts of mixture,
and § (l+x) = number of quarts of alcohol in the mixture.
Since no alcohol has been added, this must equal the number
of quarts of alcohol in the mixture at the beginning.
Multiplying both sides through by 100 we get:
50 (l+x) =95
50+50* =95
50x =45
/ _ 45 _ 9
x — S1J — TO 
.'. ^ quarts of water must be added.
88 MATHEMATICS FOR TECHNICAL SCHOOLS
3. The sum of $1100 is invested, part at 5% and part at 6%
per annum. If the total income is $59, how much was invested
at each rate?
Let $x = amount invested at 5%
then income = T % li $x
and $(1100— x) = amount invested at 6%
then income = T f 7y $(1100— x)
ordfos+dhr (1100x)=59.
Multiplying through by 100 we get:
5x +6600 6x = 5900
6600 5900 = 6x5x
700 = a:
.*. $700 invested at 5% and $400 at 6%.
Exercises XLVII.
Solve the following and verify :
1. 3x+x = 64. 17. 2x =4.
2. 5x+4x = 81. x X
3. 8x3x = 50. 3~2*
4. 13x3x = 100. Z^_oi
5. 4x+7 = 3x + 10. iy  9 " ■
6. 9x6 = 7x4. 2Q 3 = _x_
7. 7x+3=3x+67. * 4 12*
8. 2x7 = ll4x. 21 H^_19f =
0. 273x = 684x. ' 13 31
10. 423x=489x. 22 *3
11. 6x18 = 4x83x+5. ' 5
12. 10x106x27 = 3. „, 2xl
13. 24x + 1020x + 100 = ' 3
5x+96. 3x+5 =
14. 5x=05. ' 7
15. 8x=24. 25. f (x10)=0.
16. 7x = 21. 26. i (6x15)=0.
SIMPLE EQUATIONS 89
27. 3 (3z + l)(zl)=6 O + 10).
28. 3 (2a: + 5)(4x12)=5 (3x + l)4.
29. (lLr22)(86x)(48x)=17a:+7.
30. z(x+4)=x 2 +36.
31. x 2 +2:r=z 2 +4.
32. 3z 2 5(3x 2 a:)=0.
33. f +i;+.
4 5 4 o 2
«u g+3 y+5_g+9 x+4 13
4 + 2 8 + 3 "^12*
q* *+ 3 « ,x + l_a+5 1
__ x x + 2 3+x
36  3 +_ T = ^
37. ^+3* *~+2*+9.
38. 09x01x=1406a:.
39. 03:r+02=17.
40. • 007a: • 008 =004x+ 412.
' 25x+025 2z+45 . a
42 ' , 125 = T^ +  6 
Exercises XL VIII.
1. A tree 84 ft. high is broken so that the length of the part
broken off is five times the length of the part standing. What
is the length of each part?
2. After selling \ of his farm and then \ of what was left a
man still has 140 acres. How many acres had he at first?
3. The length of a rectangular building is b ft., the width is
70 ft., and its area is 43,400 sq. ft. Find the value of b.
4. A rectangular building is 84 ft. long and x ft. wide. Find
x if the area is 13,440 sq. ft.
5. A rectangular shop is x ft. long and y ft. wide. If x = 120
and y = 48 what is the area?
90 MATHEMATICS FOR TECHNICAL SCHOOLS
6. The desired area of a new rectangular boiler shop is
x sq. ft. Owing to the space available the width is limited to
b ft. What must be the length c, if 6 = 64 and x = 9648?
7. The length of a rectangular machine shop is x ft., the
width 50 ft., and the floor space must be capable of accommo
dating 20 machines, each occupying an average of 300 sq. ft.
Find the value of x.
8. The front section of an engine frame is required to
have 60 sq. in. area, the width is 5 in., and the depth is x in.
Find x.
9. A man saves $100 more than  of his salary, spends 4
times as much for living expenses as he saves, and pays the re
mainder which is $500 for rent. What is his salary?
10. If air is a mixture of 4 parts of nitrogen to 1 part of
oxygen, how many cubic feet of each are there in a room 20
ft. by 30 ft. by 10 ft.?
11. The length of a room is to its width as 4 is to 3 and its.
perimeter is 70 ft. Find the width of the room.
12. The numberplate on an automobile has a perimeter of
48 in., and its length is to its width as 3 is to 1. Find its length
and width.
13. Sirloin steak costs 1 times as much as round steak.
Find the cost per lb. of each if 3 lb. sirloin and 5 lb. round steak
cost $3.04.
14. If 2 lb. butter cost as much as 5 lb. lard, and 4 lb. lard
and 6 lb. butter cost $5.07, find the cost of each per pound.
15. The interest on $138 for a certain time at 6% per annum
is $16.56. Find the time.
16. A can do a piece of work in 6 days, B can do the same
work in 8 days, and C in 24 days. In how many days can they
do the work if they all work together?
17. A tank is emptied by two pipes; one can empty the tank
in 30 min., the other in 25 min. If the tank is f full and both
pipes are opened, in what time will it be emptied?
18. Four pipes discharge into a cistern; one fills it in one day,
the second in two days, the third in three days, the fourth
in four days. If all run together bow soon will they fill the
cistern?
SIMPLE EQUATIONS 91
19. A train runs 100 miles in the same time as a second
train runs 120 miles. If the rate of the first train is 5 miles
an hour less than that of the second train, find the rate of
each.
20. How many quarts of water must be mixed with 250
quarts of alcohol 80% pure to make a mixture 75% pure? (No
allowance for contraction).
21. A man sells \ his interest in a factory and later sells
\ of what he has left. His interest is then worth $75,000.00.
How much was his original interest worth?
22. A lot of brass scrap weighing 500 lb. contains 25% zinc.
How many pounds of zinc must be added in melting to increase
the percentage of zinc to 34%?
23. From a tank onehalf full of crude oil, 500 gallons are
drawn and 25 gallons are lost by evaporation and leakage.
If the tank is then onequarter full, how much does it hold
when full?
24. The sales of a firm increased 10% the second year
over the first, and the third year they were 20% more than
they were the second year. If the sales total $235,125.00
the third year, how much were they for the first year?
25. What is the value of the property of a person whose
income is $645.00, when he has twothirds of it .invested at
4%, onefourth at 3%, and the remainder at 2%?
26. If a boy weighing 75 lb. sits 6 ft. from the fulcrum,
where should a boy weighing 100 lb. sit to balance the beam?
27. A weight of 200 grams is placed 25 centimetres from
the fulcrum. How far from the fulcrum must a weight of
onehalf a kilogram be placed to balance the beam?
CHAPTER VIII.
FUNDAMENTAL OPERATIONS.
77. Addition. In a previous section we dealt with the
addition of simple expressions such as 6a and 3a, 4x 2 and
3x 2 , etc. We now wish to deal with compound expressions
such as 2a+56, 6a — 4x+36 2 , etc. If we wish to add a number
of these compound expressions we must recall what was
stated with respect to like and unlike terms. It was there
pointed out that like terms may be added, as for example,
6a and 2a, giving 8a. It was also stated that unlike terms
could not be added in the above way but merely written
with a plus sign between them. Thus 6a 2 plus 76 2 would be
written 6a 2 +76 2 . If then we wish to add 3a — 56+c,
2a +46 — c and 66+ 7a — 2c greater accuracy may be secured
by arranging so that the like terms would be in the same
vertical columns.
Thus, Example 1: 3a — 56+c
2a+46c
7a+66+2c
Sum = 12a +56+ 2c'
Example 2: Add 5ax — 7by\cz, ax\2by—cz,
Sax + 2by+3cz.
Arrange as above giving :
5ax — 7by+cz
ax\2by — cz
— 3aar+26i/+3cz
3ax — 36y+3cz
92
FUNDAMENTAL OPERATIONS 93
Exercises XLIX.
Add:
1. x 3 3x 2 , Sx 2 4x, 4z + l.
2. 3(xl), 4 01).
3. x — 2y+3z, 2x\y — 3z, x — 2y\z.
4. a+6, a — 6.
_ a b a.b
5 * 2 + 2' 2 + 2*
6. a — c, b—c.
7. z 2 2:n/+2/ 2 , x 2 +2xy+y\
8. x+y — 2, 3x — 2y+4z.
9. z(?/+z), 2/(xz).
10. 4(z*/), o(xy), Q(xy).
Find the values of the following sums when x = \, y = \, z = \,
a = 3, 6 = 2, c = \.
11. §af£&— c, a — j6 — §c, oa — 6+2c.
12. 5xy — 5x 2 y — 5xy, \xy\^x 2 y.
13. faffc+fc, fai&+c.
14. 12j/z8x?/+ia+6c.
78. Subtraction. In its most elementary form subtraction
has already been dealt with in connection with like terms.
Thus, 6a — 2a = 4a.
7a — 9a = —2a.
Also the rules for the removal of brackets would deal with
an expression such as 6a — (—3a). We could write this
expression 6a — (0 — 3a) =6a — 0+3a = 6a+3a.
also, — 7x — ( — ox) = — 7x — (0 — 5x) = — 7x — 0+5a;
= — 7x + 5.r.
An examination of the operation and the result in the two
latter examples brings us to a very important result with
respect to subtraction. In the first example we see that the
subtracting of —3a from 6a is equivalent to adding f3a to
6a; in the second that the subtracting of — ox from — 7.c is
the same as adding +ox to — 7x. This gives us the funda
mental principle with respect to subtraction: — To subtract one
94 MATHEMATICS FOR TECHNICAL SCHOOLS
expression from another we change the sign of the quantity to be
subtracted and add it to the other expression.
An examination of the following examples in subtraction
placed as in Arithmetic would illustrate this:
6 4a 7x 2 Sab 6x 2
9 2a 8x 2 2ab 3s 2
3 +2a x 2 +5a6 3x 2 .
If we wish to subtract one compound expression from
another we arrange as in addition. Thus to subtract 3a — 26fc
from 46 — 6a — 3c we write
46 6a 3c
26+3a+c
66 9a 4c.
Exercises L.
Subtract:
1. 4a36+c from 2a36+c.
2. a36+5c from 3a66+2c.
3. 2x — Sy\z from 15?/ — 6x+4z.
4. — 4:xy{2yz — lOzx from Zxy—6yz\7zy.
5. 4x 2 6x+2 from 7z 2 3x4.
6. From the sum of 3a+26 and 7a — 36 subtract 3a — 6.
7. Subtract 5x 2 +3x — 1 from 6x 3 and add the result to
3z 2 +2x + l.
8. Add the sum of 2y — Zy 2 and 1— 4y 3 to the remainder
obtained when l—fy 2J t2y is subtracted from 8?/ 3 +3.
79. Multiplication. The method of representing the pro
duct of two simple expressions has already been given, thus
the product of a and b = ab, the product of a, 6, and c = abc,
the product of x, y, z, and k=xyzk.
Combining this with our index laws we can find the product
of expressions like x 2 y 2 and xy giving x 2 y 2 Xxy = x 3 y 3 .
Also, 3x 2 X7x 2 = 3X7 Xx 2 Xx 2 = 21x*
and, 4x 3 X 2 = 4X2Xz 3 X, = 8a; 3  2 = 8x.
FUNDAMENTAL OPERATIONS 95
In the section dealing with brackets it was seen that
3(a+6) = 3a+36. In this case one of the expressions, 3, is a
simple expression while the other a +6 is a compound ex
pression.
If now we wish to multiply two compound expressions
together, say x\a by x\b, we may write it in the form
(x+a)(x+b).
The work may be conveniently arranged thus,
x + 6
x\a
x 2 \bx
\ax\ab
x 2 \bx\ax{ab.
Multiply x +6 by x f then multiply x+b by a and add
the results.
Example: Multiply x+2 by x+3
x+2
x+3
x 2 +2x
+3s+6
z 2 +5a:+6.
80. Rule of Signs in Multiplication. In the examples
given above all the signs are plus. It is necessary to consider
cases where the signs are minus, or some plus and some minus.
We might first recall the meaning of multiplication as
understood in Arithmetic. The fundamental unit was + 1
and all numbers were obtained from this unit.
Thus, 3 = 1 + 1 + 1.
Also, 3X4 = 3+3+3+3.
96 MATHEMATICS FOR TECHNICAL SCHOOLS
From this multiplication might have been denned as
follows: — To multiply one number by a second is to do to the
first what was done to unity to obtain the second.
This law applies with equal force to the multiplication of
fractions. Thus to multiply f by f we do to f what was done
to unity to get f : that is, we divide f into four equal parts
and take three of them. Each part would be ^j, and
by taking three of these parts we get £ X 3 =  X f.
We will, therefore, make the above definition the basis of
the rule of signs in multiplication.
(1) To multiply+3 by+4,
+3X+4 = +3+3+3+3 = +12,
or generally +aX+&=+a&.
*(2) To multiply 3 by +4.
If we do to — 3 what was done to unity to obtain 4 we
have 3 X +4 =3 3 3 3 =12,
or generally — a X +6= —ab.
(3) To multiply+3 by 4.
To obtain —4 from the fundamental unit we changed its
sign and took it four times. If this be done with +3 then
+3 X 4 =3 3 3 3 =12,
or generally +aX — b= — ab.
(4) To multiply 3 by 4.
Explaining —4 as in (3) and applying definition we have
3X 4= +3+3+3+3 = +12,
or generally — a X — b= \ab.
The results of (1), (2), (3), (4) may be stated in words
giving the following rule for signs in multiplication: — The
product of two numbers with like signs is positive and with
unlike signs is negative.
FUNDAMENTAL OPERATIONS 97
Exercises LI.
Multiply:
1. 3a by 2. 7. 7x 3 by3z. 13. x 2 y 2 z 2 byxyz.
2. Sx by 2. 8. a 2 b by ab. 14. fcrbyfy.
3. 26 by 4. 9. 4a: 2 by 2x. 15. fa 2 by 6 3 .
4. 3a 2 by a 2 . 10. p 3 byp 2 . 16. §x 3 byfx 2 .
5. 3a6 by 2a6. 11. a 3 6 bya6 3 . 17.±x 2 yby&xy 2 .
6. Zxbyly. 12. p 11 by p 3 . 18.  1 3 T a6 2 by^a 2 6.
19. \x 2 y 2 by ^. 20. 4:X 2 y by5x 3 y.
x y
Write down the continued product of:
21. 3,4,6. 25. 2a, 36, a. 29. x, x,x, x.
22. a, 6, c. 26. 2x, dx, 4z. 30. 3p 2 , 2pq, iqp.
23. a 2 , 6 2 , c 2 . 27. a 2 x, x, 2/. 31. 2x,Sx 2 ,2x i ,x 5 .
24. 6 2 ,c 2 , a. 28. 2x,2x,2x. 32. a 2 , 6 3 , 2c.
Write down the values of:
33. (x) 3 . 39. (2xy) 3 . 45. (a; 3 ) 5 .
34. (a) 4 . 40. (1) 2 . 46. (2a 2 6) 2 .
35. (2a) 3 . 41. (1) 3 . 47. (3x 2 y) 3 .
36. (x 2 ) 3 . 42. (1) 4 . 48. (3x 2 2/) 4 .
37. (a) 6 . 43. (1) 5 . • 49. \7x 2 y 2 ) 2 .
38. (x 2 ) 3 . 44. (X 2 ) 7 . 50. lxyz) 3 .
Exercises LII.
Multiply:
1. a+6— c by 4. 6. a 2 — a6+6 2 by — a.
2. 2a36+c by 2. 7. 3z 4 2;r 3 f6 by5z.
3. x+^+2 by 2i. 8. 3a 2 2a6+6 2 by26 2 .
4. 3x 2 +y 2 by2z. 9. l2x\x 2 by2x.
5. x 2 \2xy\y 2 by x. 10. x 2 —y 2 by—xy.
Find the continued product of:
11. a+6, a, 6. 14. a — 6, a, —6.
12. a 2 2a6+6 2 , a, 6. 15. z 4 3x 3 +2x 2 l,3x,2x.
13. x 2 5z+3, x 2 , x. 16. a 3 a 2 6+a6 2 6 3 , a, 6.
98 MATHEMATICS FOR TECHNICAL SCHOOLS
When a = — 2, 6= —3, find the value of:
17. a 2 2. 22. 6 4 81. 27. a 2 +66 2 .
18. 2a 2 a+2. 23. 6 2 a 2 +2a. 28. a 4 b\
19. a 2 b\ 24. a 3 + 8. 29. a 5 6 5 .
20. a 2 2ao+6 2 . 25. a 3 +b\ 30. a 3 36.
21. 2a 3 + 16. 26. 8a 2 6 3 . 31. a 4 l.
Exercises LIII.
Find the product of :
1. z + a, x — b. 11. x 2 — a 2 ,xAa.
2. ay 6, q/d. 12. x + 2y, 3x + l.
3. 5+3x, 72*. 13. 7a2b, a 2 b 2 .
4. x — by, 2x+?>y. 14. ax 2 — bx, ax\b.
5. af3ar, a — ox. 15. 6a — 26, a — 6.
6. az + 1, 6x + l. 16. a+6, c— d.
7. 4a 2 36, 2a 2 6. 17. x 2 +a, x 3 b.
8. a: 3 — 1, x + 1. 18. bx — ay,ax—cy.
9. a 2 +66, a 2 46. 19. ary»l,«y+2.
10. a+3x, abx. 20. x 5 l, x 4 +l.
81. There is a number of types of products in which the
results can be written down by inspection if a few typical
examples are examined.
(1) (x\a)(x — a) =x \a
x — a
x 2 \ax
— ax — a 2
x 2 —a 2 .
That is, the product of the sum and difference of two quantities
is equal to the difference of their squares.
Thus, (x+3)(s3) = x 2 9.
(a+b)(ab)=a 2 b 2 .
(xy + l)(xy — l) =x 2 y 2 — l.
(2) (a+6)(a+6) or (a+b) 2 = a +6
a +6
a 2 +ab
+ab + b*
a 2 +2ab+b\
FUNDAMENTAL OPERATIONS 99
That is (a+6) 2 = the square of a, plus the square of 6,
plus twice the product of a and b. Any expression consisting
of two terms is called a binomial, so that we may state as
a general rule: — The square of a binomial is equal to the sum
of the squares of the terms plus twice their product.
Thus, (x+3) 2 = x 2 +9+6x.
(x4:) 2 = x 2 +lQ8x.
(xy — iy = x 2 y 2 \\— 2xy.
Also, (a+6+c) 2 ={a+(6+c)} 2 = a 2 +(6+c) 2 +2a(6+c).
= a 2 +6 2 +c 2 +26c+2a6+2ac.
= a 2 +6 2 +c 2 +2a&+2ac + 26c.
This method may be used for the square of an expression
containing any number of terms so that the rule may be
given thus : — The square of an expression consisting of any
number of terms is equal to the sum of the squares of each of
the terms plus twice the product of each term multiplied by each
of the terms that follow it.
(3) (x +2) +3)= x +2
x+3
x 2 +2x
+3x+6
z 2 +5x+6.
Here we observe that the first term in the product x* is
obtained by multiplying the first terms in each of the factors,
the second term 5x is obtained by adding the 3 and the 2
and multiplying by x, the third term is obtained by multiplying
the 2 and the 3 together.
Thus, (x+4)(a:+5)=x 2 +9^+20.
(x4)(;rf3)=.r 2 x12.
(x6)(x4)=x 2 10xr24.
100 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises LIV.
Write down the results of the following:
1. (c+d)(cd). 14. (2x+3y) 2 .
2. (2s +3) (2s 3). 15. (xy + l)\
3. (x 2 2a 2 )(x 2 +2a 2 ). 16. (x 2 l) 2 .
4. (s 2 +2)(x 2 2). 17. (a+6c) 2 .
5. (a+36)(a36). 18. (2a6c) 2 .
6. (px+q)(pxq). 19. ( a +b+cd)\
7. (x 2 3y 2 )(x 2 +Sy*). 20. (2a36+c) 2 .
8. (2xSy)(2x+dy). 21. (z+3)(s+4).
9. (a 2 46)(a 2 +46). 22. (a+5)(a2).
10. (*+*/) Or ?/)(:r 2 +2/ 2 ). 23. (z+8)(z5).
11. (c+d)*. 24. (p+3g)(p6g).
12. (a26) 2 . 25. (a&+4)(a&5).
13. (2x^) 2 . 26. (*y6)(*y+c).
Use the rule for the square of a binomial to find the
value of:
27. 99 2 . 29. 105 2 . 31. (1006) 2 .
28. 102 2 . 30. 95 2 . 32. (995) 2 .
82. Division. The Rule of Signs in Division may be
readily deduced from the rule in Multiplication.
Thus, (1) +xy=+xx+y .'. + X y++x=+yor^p! = +y.
(2) xy= xx+y .*.. xy+x=+y or —^=+y.
— x
~\~xy
(3) +xy= —vxy .*. +xy¥ —x= — y or — f~ — y.
(4) — z?/ = +sx2/ .". —xy?+x=—y or —^=y.
+x
From these results we have the following rule of signs in
division: — Terms with like signs when divided give plus (+).
Terms with unlike signs when divided give minus (— ).
FUNDAMENTAL OPERATIONS 101
Examples: qs= +3. T7o = ~'  ^'
+6 3 ^3
±2la^ = _ 7a Zx 2 y 2 ^ 3
— 3a +xy
35a 3 6 2 c _ ,, + 5x 7 .
= 5a 2 o. — ^i=— a: 5 .
— labc — 5x 2
Exercises LV.
Divide:
1. 3xby 3. 8. 6 4 by 6.
2. 3x by 3. 9. 8a 2 by 4a 2 ,
3. 3xby3. 10. 54a 2 6c by 6abc.
4. 3xbyx. 11. 24a 2 6 2 c 2 by406c.
5. 6xy by 6x. 12. 21xV by7x 3 ?/ 2 .
6. a 2 bya 2 . 13. 49a 3 6 3 by 7a 2 6 2 .
7. 8a 2 by 4a. 14. x 5 by+x 2 .
Simplify:
15. A^ 18. 24y2 * 2 , 21 121 * V
5 —4?/ llx 3 i/
— 21x 3 ?/ 3 49pa 2 r 16a 3 6 3
— 3xi/ ' ' —7pqr' — 8a 2 b
1? 8x^ 2Q 32Z. 2 m 2 r^ ^ \abc
■xy ' 4:1m ' \abc 2 '
Divide:
24. 3x6?/ by 3. 30. 6a 96 + 12c by 3.
25. 3x9 by 3. 31. x 3 +3x 2 3x by x.
26. 3x 2 6x by 3x. 32. 15?/ 4 5?/ 3 x 3 30?/ 3 by by.
27. 6 2 +a6by6. 33. 5m 3 n+20m 2 w 3 by 5mn.
28. 4a 2 68a6 2 by 2a6. 34. a 2 bcab 2 c\abc 2 by abc.
29. x 3 +x 2 byx 2 . 35.  a 2 b 2 c 2 + abc 2  cab 2 by abc.
102 MATHEMATICS FOR TECHNICAL SCHOOLS
83. To divide one compound expression by another the
work may be arranged by following the method of long
division in Arithmetic:
Example. Divide x 2 +5x+G by x+2.
s+2)x 2 +5x+6 /x+3.
x 2 +2x (1)
3x+6 (2)
3s+6. (3)
x 2 ix = x .'. x is the first term in the quotient, (x+2) multi
plied by x gives x 2 \2x and we obtain (1). Line (2) is obtained
by subtracting x 2 + 2x from the expression and bringing
down +6. 3x divided by x = 3, .*. 3 is the second term of the
quotient.
(x+2) multiplied by 3=3x}6 and we obtain line (3).
This when subtracted leaves no remainder and the quotient
is x+3.
This method may be applied to an expression of any number
of terms, if care is taken to arrange the divisor and dividend
in descending or ascending powers of some common letter,
and to keep the remainder in each case in the same order.
Exercises LVI.
Divide:
1. x 2 +7x + 12byx+3. 9. 2530a+9a 2 by 53a.
2. a 2 +3a+2 by a+2. 10. 4x 4 49 by 2x 2 7.
3. a 2 — 3a+2 by a — 1. 11. x 2 \ax\bx\ab by xfa.
4. x 2 — 5x — 14 by x+2. 12. x 4 \x 2 y 2 \y 4 by x 2 — xy+y l .
5. 15x 2  26x48 by 5x2. 13. a 3 +6 3 by a+6.
6. 613a+6a 2 by 23a. 14. x b +bx*y + l0x z y 2 + \0x 2 y* +
7. 4+4x+x 2 by 2+x. 5xy A +y 5 by x 2 +2xy+y 2 .
8. x 2 \2xy+y 2 by x+y. 15. a 3 +6 3 +c 3 — 3a6cby a + 6+c.
CHAPTER IX.
FORMULAS.
84. One of the most Valuable Uses for algebraic symbols is
to express a scientific law in a short form. When such a law
is expressed in algebraic form it is called a formula. For
example, the area of a rectangle is equal to the length multiplied
by the breadth. If we let A represent the area, I the length,
and b the breadth, we could briefly represent this relation by
the equation A=lb.
If Z = 15 in., 6 = 10 in., then A = 15 X 10 = 150 sq. in.
Again if A =lb., then l = i This is called solving for I.
If ,4=200 sq. in., 6 = 25 in., then Z=^=8in.
Further if A— lb., 6 = 7 This is called solving for 6.
If ^1=400 sq. in., Z = 40in., then 6=^00. = 10 in.
If a scientific law be stated in detail it is important to be
able to express it as a formula.
Example.
To find the number of revolutions of a driven pulley in a
given time, multiply the diameter of the driving pulley by its
number of revolutions in the given time, and divide by the
diameter of the driven pulley.
Using D and d for the diameters, and N and n for the number
of revolutions respectively, express the above as a formula.
Exercises LVII.
1. The cutting or surface speed, that is the number of linear
feet measured on the surface of the work that passes the edge
of a cutting tool in a minute, is found by multiplying the
circumference of the work being turned by its R.P.M. Express
the rule as a formula.
103
104 MATHEMATICS FOR TECHNICAL SCHOOLS
2. The current flowing along a conductor is given by the
E
formula 1= „, where / is the current in amperes, E the electro
it
motive force in volts, and # the resistance in ohms.
Solve for E and #.
If £=110, # = 220, find/.
If/=2, £ = 220, find R.
If/ =5, R= 75, find E.
3. The resistance of a wire in an electric circuit is given by
R=K j, where R is the resistance, L the length of the wire,
A its area in circular mils, K the resistance of 1 mil foot in ohms.
Solve for K, L and A.
IfZ = 3ft., ,4 = 1000, # = 105, find R.
If # = 220, K = 105, L = 1000 ft., find A.
It R= 10, # = 105, ,4=250, find L.
4. The work done by any force is given by W = FS, where W
is the work done, F the force in pounds, S the distance in feet
through which the force acts.
Solve for F and S.
If F = 525 lb., S = 51ft., find W.
If FT = 150, F = 5oz., findS.
If W = 500, S = 800 ft., find F.
5. The total resistance of a series circuit is given by
R = R l \R 2 {R 3 , where R is the total resistance and R 1} # 2 , # 3
are the resistances of the separate parts of the circuit re
spectively.
Solve for R lf R 2 and R 3 .
tt R,= 2, #2 = 4 , #3 = 9, find R.
If #=12, #]=2, # 2 = 7, find # 3 .
6. The indicated horsepower of a single acting engine is
given by:
PLAN
33000'
where P is the mean effective pressure on the piston in pounds
per sq. in., L the length of the stroke in feet, A the effective
area of piston in sq. in., N the number of strokes per minute.
FORMULAS
10E
Solve for P,
L, A and N. •
If P =80,
Z = 2ft., ^4=30sq.
in.,
N= 60,
find 77. P.
Ifff.P.= 4,
Z = Uft., ^l=24sq.
in.,
N= 50,
find P.
If ff.P. = 10,
p = 50, ^l=30sq.
in.,
iV=100,
find Z.
If #.P.=20,
p = 60, Z = 2 ft.,
# = 100,
find A.
If fl.P. = 16,
P = 60, i = 2 ft.,
A= 40,
find iV.
7. The formula D=QIT is used in electrolysis, where D is
the weight of the deposit, Q the electrochemical equivalent,
T the time in seconds, 7 the current.
Solve for Q, I and T.
If Q = 001118, 7=40, T = 600, find D.
If 7)= 2, Q= 000328, 7 = 250, find T.
If7)=5, 7=10, Z" = 153, find Q.
8. The diameter of a rivet is given by d = 1 • 2^/t, where d
is the diameter in inches and t the thickness of the plate in
inches.
lit =75, find d.
If d = £g in., find t.
9. The space through which a body falls from rest is given
by s = h 9t 2 > where s is the space in ft., g the acceleration due
to gravity, t the time in seconds.
Solve for g and t.
If * = 12, g =322, find *.
If 5 = 3155.6, = 322, find t.
W
10. In a machine E= 5p,, where 7? is the efficiency, W the
weight, V the velocity ratio, P the horizontal force.
Solve for W, P and F.
IfJF = 112, P = 20, F= 12 5, find E.
If£=55, P = 25, F=185, find W.
If E = • 74, IF = 350, K = 23, find P.
If £=346, JF = 79926, P = 20, find V.
11. The allowable working pressure in a steam boiler is
givenby: „ 2Tsk
DF '
*
where T is the thickness of the plate in inches, s the tensile
strength of plate in pounds per sq. in., k the efficiency of the
106 MATHEMATICS FOR TECHNICAL SCHOOLS
joint, D the inside diameter of shell in in., F the factor of
safety.
Solve for T, s, k, D and F.
If T= h 5 = 35000, fc= 45, D = 30, F = 4, find B.
IiB = 75, s = 40000, k=5, D = 40, F = 5, find T.
If 5= 38, r = A, fc«5, D = 50, P = 4, find 5.
If 5 = 150, r= , * = 60000, Z) = 60, P = 5, find A:.
If 5 = 200, r1, 5 = 70000, &=75,F = 5, find Z).
If 5= 40, T=i, 5 = 65000, fc=8, Z) = 120,findF.
12. The horsepower of an electric current is given by
EI
H.P. = =j~, where E is the electromotive force and 7 the
current in amperes.
Solve for E and 7.
If E =110, 7 = 30, find H.P.
If H.P. = 6, E = 200, find /.
If H.P. = 10, 7 = 40, findE.
13. The heat generated by a current is given by H = • 24 EI T
(Joule's Law) where H is the heat in calories, E the electro
motive force, I the current in amperes, T the time in seconds.
Solve for E, I and T.
If £ = 110, 1=2, T= 30, find H.
If # = 500, E= 6, I = 10, find Z\
If # = 1000, 7 = • 5, T = 160, find £.
14. The space traversed by a body starting from rest and
moving with a uniform velocity is given by s = vt, where s is
the space, v the velocity and t the time.
Solve for v and t.
If v = 12 ft. per sec, < = 25 sec, find s.
If 5 = 300 ft., t = 15 sec, find v.
If 5 = 500 ft., v = 20 ft. per sec, find t.
15. The width of a single belt to transmit a given horse
power is given by W= — D ^ „ — , where W is the width of
the belt in in., H the horsepower transmitted, P the allowable
pull per in. of width of belt, S the speed of the belt in feet
per min.
FORMULAS 107
Solve for H , P and S.
If W = 13, P = 30, S = 3000, find W.
IiW= S, P = 40, S=3500, find H.
UW= 6, # = 20, S = 3200, find P.
UW= 6, H = 22, P= 30, findS.
16. The brake horsepower ' of an engine is given by
0_ p D \7
B. H.P.= , where B.H.P. is the brake horsepower, P
ooUUU
the reading of the scale beam, P the length of the arm
in ft., N the revolutions per min.
Solve for P, R and N.
If P =180 lb., P= 48 in., N = 250, find B. H. P.
IfP.ff.P.=30, R= 54 in., iV=120, find P.
IfP.#.P. = 32, P = 2001b., #=180, find P.
IfP.ff.P.=36, P = 2201b., iJ = 5 ft., find N.
CHAPTER X.
MENSURATION OF AREAS.
85. Mensuration is that part of Mathematics which deals
with the length of lines, the areas of surfaces, and the volumes
of solids.
86. To Find the Area of a Rectangle or of a Square.
In Figure 27, ABCD is
a rectangle, i.e., a quadri
lateral with its opposite
sides parallel and its angles
right angles. If we divide
each side into inches and
join as above we see by
actually counting the small
squares that the area of
the rectangle is six square
inches (to scale). This
result might have been obtained by multiplying the number
of inches in the length (3) by the number of inches in the
width (2). From this example we infer a formula for the area
of a rectangle. If A represents the area, b the length,
and h the breadth, then A=bh, or the area of a rectangle
= length X breadth.
Note. — A correct statement of the above formula would
manifestly be — the measure of the area of the rectangle = the
measure of the length multiplied by the measure of the breadth,
but for the sake of brevity the word " measure " will be
omitted throughout.
Make drawings in your laboratory book to test the accuracy
of the above.
108
D
Fig. 27
MENSURATION OF AREAS
87. To Find the Area of a Parallelogram.
109
Fig. 28
In Figure 28, ABCD is a parallelogram ( gm ), i.e., a quadri
lateral with its opposite sides parallel.
If the rightangled triangle DFC be cut out and placed on
EBA, it will coincide with EBA. The s m ABCD is therefore
equal in area to the rectangle EBCF. If the area of the  gm
is A, the base b, and the perpendicular height or altitude h,
then A = bh, or the area of a parallelogram = base X perpen
dicular height.
Make drawings in your laboratory book to test the accuracy
of the above.
88. To Find the Area of a Triangle in Terms of its Base and
Altitude.
Fig. 29
In Figure 29, ABC is a triangle. If we draw CD parallel
to AB and AD parallel to BC, we have the « m ABCD. Since
no
MATHEMATICS FOR TECHNICAL SCHOOLS
the area of the  gm A BCD is bisected by its diagonal AC, we
have the area of the triangle ABC as onehalf the area of the
 gm ABCD. If A is the area of the triangle, b its base, and
h its altitude, then A=%bh, or the area of a triangle = \ base
X altitude.
Make drawings in your laboratory book to test the accuracy
of the above.
89. To Find the Area of a Triangle in Terms of the Sides.
In Figure 30 we have a
triangle ABC and have de
noted the sides by a, b, c; a
being opposite angle A, b
opposite angle B, and c
opposite angle C.
The area of the triangle
is given by the formula:
A = \/s(sa) (sb) (sc) where
a, b, c are the sides, and s
is onehalf their sum.
Example: — If the sides in Figure 30 are 13 ft v 14 ft., 15 ft.
respectively, then 5 = 21, sa = 8, sb = 7, sc = Q.
.*. .4 = V2lX8X7X6 = \/7056 = 84sq. ft.
Exercises LVIII.
1. Supply the missing quantities in the following rectangles :
Area Length Breadth
sq. ft.
4 ft.
3 ft.
444 sq. ft.
37 ft.
ft.
3605 sq.ft.
ft.
18 9 ft.
sq. yd.
24 ft. 9 in.
15 ft. 6 in.
f acre
ft.
2\ chains
MENSURATION OF AREAS
111
2. Supply the missing quantities in the following parallelo
grams:
Area
Base
Altitude
sq. ft.
4 ft.
3 ft.
48400 sq. yd.
352 yd.
yd.
sq. ft.
2 ft, 3 in.
8 in.
378 sq. in.
3 ft. 6 in.
in.
sq. yds.
5 yd. 1 ft.
3 yd. 2 ft.
90. To Find the Area of a Trapezium.
Fig. 31
Figure 31 represents a trapezium, i.e., a quadrilateral with a
pair of sides parallel.
The diagonal AC divides the trapezium into the two triangles
ABC and ADC.
Area of ABC = \ yh.
Area of ADC = \ xh.
/. A = \ yh+%xh = \h(x+y),
or the area of a trapezium = sum of parallel sides X h the per
pendicular distance between them.
112
MATHEMATICS FOR TECHNICAL SCHOOLS
91. A Practical Application of the Triangle and the Trapezium
is found in the Measurement of Land.
The following represents an entry in a surveyor's field book
and the corresponding plan:
Plan
Field Book
Links
To B
460
to E 120
340
180
to D80
100
From
A
90 to C
go North.
The field book entry is read upwards,
which in the case above indicates that
the chain line runs north from A. The
centre column refers to measurement
from A along the chain line to points
from which the offsets are taken. Offsets to
indicated on the right and offsets to the left
on the left.
In the plan the area of ADX = §X100X80
the area of EDXZ=\ X240( 120 +80) «
the area of BEZ = ^X120X120
the area of A YC =1X180X90
the area of B YC = §X280X90
the right are
are indicated
= 4000 sq. li.
= 24000 sq. li.
= 7200 sq. li.
= 8100 sq. li.
= 12600 sq. li.
Total Area =55900 sq. li.
= 559 acres.
MENSURATION OF AREAS
Exercises LIX.
1. Find the missing quantities in the following triangles:
113
Area
Base
Altitude
sq. in.
24 in.
13 in.
sq. ft.
2 ft. 6 in.
3 ft, 4 in.
120 sq. in.
in.
15 in.
22 sq. ft.
5 ft, 6 in.
ft.
5§ acres
320 rods
yards
2. Find the missing quantities in the dimensions of the fol
lowing boiler plates in the form of trapeziums:
Area
sq. in
6 ft.
62 in.
102 in.
sq. in.
6 ft. 9 in.
77 in.
83 in.
8505 sq. in.
11 ft. 10 in.
101 in.
in.
sq. in.
8 ft. 5 in.
79£ in.
93 in.
9841f sq. in.
98 in.
90 in.
549 sq. ft.
25 ft.
18 ft.
3. Find the areas of the following triangles:
Sides 3 ft., 4 ft., 5 ft.; answer in sq. feet.
Sides 4 yd., 2 ft., 3 yd., 2 ft., 1 yd., 1 ft.; answer in
square yards.
Sides 17 in., 18 in., 19 in.; answer in sq. inches.
114 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises LX.
1. Find the areas of the triangular faces of a number of the
models in the laboratory, using both methods. Make drawings
in your laboratory book.
2. Find the areas of trapeziums available in the laboratory.
Make drawings in your laboratory book.
3. A rhombus is a quadrilateral with all its sides equal.
Construct a rhombus in your laboratory book having each side
2 in. Employ both experiment and equality of triangles to
establish how one diagonal divides the other, and also the
magnitude of the angle contained by the diagonals. Write
out the details and derive a formula for the area of a rhombus
in terms of the diagonals.
4. Take a series of measurements in the school grounds
and enter in your laboratory book as suggested. Draw
a plan to scale from your measurements and calculate the
area.
5. What is the area of the surface of a boiler plate 3' 8"
by 1' 6"?
6. How many square pieces of zinc 6" X 6" can be cut from a
zinc plate 3' X 6'?
7. What is the value of copper in an open copper tank
measuring 4f " long, 3" wide and 2\" deep; copper weighing
12 lb. per sq. ft. and costing 40c per lb.? (No allowance being
made for laps, seams or waste).
8. The diagonals of a sheet of zinc in the form of a rhombus
are 24" and 16". Find the area of the sheet.
9. If a sheet of copper 5' X 10' weighs 500 lb., what is the
weight per sq. ft.?
10. How many sq. ft. of sheet copper will be required to
make an open rectangular tank 7' long, 3' wide, and \\' deep,
allowing 12% extra for waste?
11. Find the cost of shingling the roof in the diagram on
page 62 with shingles laid 4§ in. to the weather if material
and labour cost $14 a square, of shingles, the eaves projecting
2' (equivalent to a roof 34'X28' on plan).
12. Find the cost of putting a slate roof on the building in
the diagram on page 62, gauge 85", at $30 a square.
MENSURATION OF AREAS
115
Fig. 33
13. Find the cost of covering with 1" square sheeting the
gable ends of the building in the diagram on page 61, width
20', rise 10', material to cost $50 per M, allowing 8% for
waste.
14. Find the cost of cover
ing with 1" square sheeting
the gable ends of the roof re
presented in Fig. 33 at $52 per
M, allowing 10% for waste.
15. In the map of a district
it is found, that two of its
boundaries are approximately
parallel and equal to 13 miles
and 18 miles. If the breadth
is 8 miles find the area.
16. In the quadrilateral ABCD the diagonal AC is 62"
long, and the perpendiculars on AC from B and D are 15"
and 12" respectively, find the area of the quadrilateral in
sq. ft.
17. Construct an equilateral triangle 2" to the side. Find
its altitude.
18. Construct an equilateral triangle with the length of
the side taken at random. Denote it by x and find the altitude.
What is the relation between the altitude and half the base?
19. Find the area of a regular hexagon, if one of the sides be
3". (Divide into six equilateral triangles).
20. Find the side of an equilateral triangle equal in area to a
triangle with sides 13", 14" and 15" respectively.
21. How many sq. ft. are there in the surface of a board 18'
long, 6" wide at one end and 14" wide at the other?
22. How many sq. in. are there in a triangular plate, if one
of its sides be 15", and the perpendicular on it from the opposite
vertex be 8"?
23. If the sides of a triangular plot of ground be 26', 28'
and 30' respectively, find the length of the perpendicular from
the opposite vertex on the 30 ft. side.
24. One side of a triangular plate, containing 45 sq. in., is
8". Find the length of the perpendicular on this side from the
opposite vertex.
116
MATHEMATICS FOR TECHNICAL SCHOOLS
25. The sides of a rightangled triangle are 5", 12" and 13"
respectively. Find the areas of the equilateral triangles
described on its sides. Do these areas bear any relation to
each other?
26. A column having a
opposite arms of the cross 4 2
crossshaped section has two
long, and the other two
arms 4". The arms are \" wide. What is the area of the
section?
27. A Tshaped section has the top flange 8" long and f "
wide, the other flange measuring 4' long by f " wide. What
is the area of the T1
28. The two parallel sides of a trapezium measure 13 chains
60 links, and 6 chains 40 links; the other sides are equal, each
being 8 chains 50 links. Find the area.
29. ABCD is a quadrilateral in which the following measure
ments have been taken: AB = S0", BC = 17", CD = 25",
DA =28", the diagonal BD = 2Q". Find the area in sq. ft.
30. ABCD is a quadrilateral in which the angles ABC,
CD A are right angles, and AB = 36 chains, BC = 77 chains,
CD = 68 chains. Find the area in acres.
31. Find the area of a quadrilateral ABCD in which the
diagonal AC measures 30', and the perpendiculars on it from
B and D are 3^' and 6' respectively.
32. Draw the plan and calculate the area, in acres, of a plot
of ground from the following* notes:
Links
•
To B
530
to £75
400
240
120 to C
:o D 100
150
From
A
go North.
MENSURATION OF AREAS
117
33. Draw a plan and calculate the area, in acres, from
the following notes:
Chains
to B
245
ToF 2
1526
101
316 to E
To D24
86 '
43
15 to C
From
A
go North.
34. Draw a plan and calculate the area, in acres, from the
following notes:
Links
to B
1200
250
100
760
324
400
50
360
200
From
A
Go
N. 30° W
35. How many 6 in. sq. tiles should be supplied to cover
the courtyard shown in Figure 34, an allowance of 5% being
added to cover cutting and breakage?
118
MATHEMATICS FOR TECHNICAL SCHOOLS
36. Figure 35 shows a gussetplate for a girder. What is its
weight if the plate is of mild steel \" thick, weighing 204 lb.
per sq. ft.?
t/fl
WlU'^9
U74//VJ
Fig. 34
Fig. 35
37. Calculate the length of the rafters on each pitch and the
total area of the entire gable end of the building in Figure 36.
■xf
Fig. 36
/5* —
Fig. 37
38. Determine the area of the crosssection in Figure 37.
Fig. 38
92. The Circle. A circle is a plane figure bounded by a line
called the circumference and such that every point on it is
equidistant from the centre.
MENSURATION OF AREAS
119
93. To Find the Circumference of a Circle, the following
measurements of a series of circular models were made and
tabulated as follows:
Circumference
Diameter
Circumference
— ^ = w
Diameter
1178
in.
375
in.
31413
654
in.
208
in.
3.1442
. 1098
in.
35
in.
3.1371
635
in.
202
in.
3 • 1435
1685
in.
536
in.
31436
515
in.
164
in.
3 • 1402
300
in.
954
in.
3 • 1446
1071
in.
341
in.
3 • 1407
864
in.
275
in.
31418
Average
31418
The value of tt has been determined to a great number of
decimal places but 31416 is a close approximation. Since the
fraction ^ =3142 when carried out to the third place, it is
commonly used as the value of tt. •
From the above experiment we infer that Circumference =
it Diameter or C = tZ). If in the formula C = n D we substitute
for diameter its value in terms of the radius, we obtain
C = 7r(2r) or C = 2:rr.
120
MATHEMATICS FOR TECHNICAL SCHOOLS
94. To Find the Area of a Circle. If we take a circular board
and divide it into sections as shown in Figures 39, 40, and place
them as in Figure 41, we practically have a rectangle whose
length is onehalf the circumference and whose width is one
half the diameter of the circle.
Fig. 39
Fig. 40
Fig. 41
Hence, to find the area of a circle we multiply onehalf the
circumference by the radius, i.e., Trrxr = irr 2 .
.*. A = irr 2 .
In Figure 41, how would you to some extent overcome the
difficulty of the length of the rectangle not being a straight line?
In the formula A = irr 2 if we write for r its value in terms of
D we get A „()'^?J^ = . 7854Z) ,
This formula for the area of a circle is commonly used by
engineers and machinists.
MENSURATION OF AREAS
121
Exercises LXI.
1. Supply the missing quantities in the following circles:
(tt = 31416)
Radius
Diameter
Circumference
Area
5 ft.
14 ft.
31416 ft,
150 sq. ft.
95. To Find the Area of a Circular Ring or Annulus.
The area of the outer circle is nR 2
and the area of the inner circle is
?rr 2 . .'. area of the ring = nR 2 —
7rr 2 = 7r ( j R2_ r 2) =1F (R+r)(Rr).
If we examine this latter formula
in relation to the figure we see that
U + r is the length of the mean
diameter AB, and that R — r is the
width of the ring. The formula
therefore, be
v{R+r)(R — r) may,
Fig. 43
circle by the arc ADB,
Fig. 42
written 7r(mean diameter) X (width
of ring). .*. area of ring = mean
circumference X width of ring.
96. To Find the Length of the Arc
of a Circle.
In Figure 43 the chord AB divides
the circumference of the circle into
two arcs ADB and AHB.
If the angle AOB, that is the
angle subtended at the centre of the
is 120° then the length of the arc
122 MATHEMATICS FOR TECHNICAL SCHOOLS
ADB is ^f# X circumference or in general the length of the
n
arc ADB is ^r^X circumference, where n is the number of
3b0
degrees in the angle subtended at the centre.
The length of an arc may be found approximately by the
formula: — Length of arc = — 5 — where c is the chord of the
arc and a is the chord of half the arc.
97. To Find the Area of a Sector of a Circle. A sector of a
circle is that part contained by two radii and the arc cut off
by them.
In Figure 43, KOH represents a sector of a circle. If the
angle K H be 60° the area of the sector will be ££$ of the
n
area of the circle, or in general the area of the sector is ttfk X
00U
area of the circle, where n is the number of degrees in the angle
contained by the two radii.
By a method similar to that used in finding the area of a
circle it may be shown that the area of the sector =% arc of
sector X radius of circle.
98. To Find the Area of the Segment of a Circle. A segment
of a circle is that part of the circle contained by an arc and its
chord.
In Figure 43 the chord A B divides the area of the circle into
two segments, the area above AB being called the minor
segment, and the area below AB the major segment. It is
evident that the area of the segment ADB is equal to the
area of the sector AOB minus the area of the triangle AOB,
so that if sufficient data is given the area may be found by this
method. The area of the minor segment in Figure 43 may be
found approximately from the formula.
Area of Segment = y +§ ch. Where c is the length of the
chord AB and h is the height CD.
MENSURATION OF AREAS 123
Exercises LXII.
1. Measure a number of circular objects as suggested on
page 119. Tabulate in your laboratory book, find the ratio
of the circumference to the diameter in each case, and take
the average of these results.
2. What length of steel sheet would be needed to roll into a
drum 42" in diameter?
3. What is the circumference in feet of an 18" emery wheel?
4. The 72 in. drivers on a locomotive make 245 turns per
min. How many feet will the locomotive go per min.? How
many miles will it travel per hour?
5. Find the diameter of a driving wheel measuring 15'
8^" around the outside?
6. If a point on the rim of a flywheel should not travel over
a mile a minute, what should be the maximum speed, in
revolutions per min., of a flywheel 7' in diameter?
7. A pulley 3' in diameter makes 200 revolutions per min.
Through how many feet does a point on its rim travel in 2
minutes?
8. If a belt connects the pulley in the preceding question
to a 15" pulley, through ho.w many feet will a point on the
rim of the latter travel in 1 minute?
9. If a speed of a mile a minute is desired, what size emery
wheel should be ordered to go on a spindle running 1320
R.P.M.?
10. A degree of latitude in Toronto measures 26461331 ft.
Find in miles the length of the parallel of latitude passing
through Toronto.
11. The perimeter of a semicircle is 72", find its radius.
12. A pulley 36" in diameter drives another pulley 14"
in diameter. The belt velocity is 22' per second. What are
the R.P.M. of the pulleys?
13. The flywheel of an engine is 12' 6" in diameter and
revolves at 96 R.P.M. It is belted to a 48" pulley on the main
line shaft. Find the speed of the shaft.
14. A 36" pulley making 143 R.P.M. is belted to another
making 396 R.P.M. Find the diameter of the latter.
15. Describe a circle of given radius on your crosssection
paper. Count the squares as accurately as possible. Con
struct a square on the radius and count the squares. Divide
124
MATHEMATICS FOR TECHNICAL SCHOOLS
the former result by the latter. Repeat this experiment
changing the radius in each case. Tabulate in your laboratory
book and find the average of your results.
16. To establish A = 7854Z) 2 experimentally, cut a circular
piece of cardboard 1' in diameter and also a square piece of
the same material 1' to the side. Weigh both and find the
value of — r^j— — j — r^— j — . Repeat this with pieces of board,
weight of circle
pieces of zinc, etc., taking care that the materials, in any one
case, have the same thickness and density, and that the
diameter of the circular part is the same as the side of the
square. Tabulate in your laboratory book and find the
average of the results.
17. Fill in the omitted entries in the following:
No.
Diameter
Circumference
Area
1
7
2
44
3
•
154
4
176
5
201
6
5544
7
148
8
264
9
154
18. The piston of a locomotive is 20" in diameter. Find its
area in sq. in. If the highest pressure carried is 205 lb. per
sq. in., what would be the total pressure tending to blow off
the cylinder head?
19. A workman finds the circumference of a shaft to be 11".
In order to find the strength of the shaft he must know the
area of a crosssection. Find this area.
MENSURATION OF AREAS 125
20. Which has the greater capacity, one 4" pipe or two
2" pipes?
21. The area of an 8" circle is how many times the area of a
4" circle? The area of a 12" circle is how many times the area
of a 4" circle?
22. Employ the relation between the sides of a rightangled
triangle to find the diameter of a pipe equal in carrying capacity
to two pipes 2" and 3" in diameter respectively. Illustrate
by means of a diagram. Extend this method of illustration
to find the diameter of a pipe equal in carrying capacity to
three pipes 2", 3", 4" respectively, in diameter.
23. The total pressure in a cylinder is to be 6000 lb. If
the pressure per sq. in. is 50 lb., what is the diameter of the
piston?
24. A circular duct in a heating system is to supply air
for four rectangular outlets 6" by 8". What must be the
diameter of the duct so that its capacity will be equal to the
combined capacity of the four outlets?
25. Establish experimentally the formula — Area of ring =
mean circumference X width — by considering the ring as a
trapezium.
26. The inner and outer diameters of a ring are 9" and
10" respectively, find the area of the ring.
27. A hollow castiron column has inside and outside
diameters of 12" and 16" respectively, find the area of the
end of the pipe.
28. What is the area of a circular race track 378 yd. inside
diameter and 16' wide?
29. What is the area of the end of a castiron pipe that is
12" outside diameter and 1" thick?
30. What is the area of the end of a rod that is 4" outside
diameter, and has a \\" hole running through the centre
of it?
31. A circular court 150 yd. in diameter is to have a walk
10' wide around it on the inside. The remainder is to be
sodded. Find the totalcost if the walk costs $2.00 a sq. yd.
and the sodding 40c a sq. yd.
32. The radius of a circle is 8'. Find the area of a sector
of the circle, the angle of which is 36°.
126
MATHEMATICS FOR TECHNICAL SCHOOLS
33. Find the radius of a circle such that the area of a sector
whose angle is 60° may be 1825 sq. in.
34. Find the area of the sector of a boiler supported by a
gussetstay, the radius of the boiler being 42" and the length
of the arc 25".
35. The centres of two circles which intersect aTe 12' apart.
The radius of the one circle is 9', and that of the other 8';
find the area of the part which is common to both circles.
36. Find the area of the segment of a circle if the chord
be 15" long and the height of the arc 6".
37. Construct an arc of a circle by tracing part way around
any circular object. Join the ends of the arc to form the
segment of a circle. Find the centre of the circle and determine
the length of the arc by treating as part of the total circum
ference. Also find the length of the arc by the formula
— r — , and hence determine the percentage error in this
formula.
38. Find the area of the segment in the above by finding
the asea of the sector and subtracting the triangle. Also,
h 3
find area by the formula — +f ch, and hence determine the
JZG
percentage error in this formula.
99. The Ellipse. An ellipse is a plane figure bounded by
a curved line, such that the sum of the distances of any point
in the bounding line from
two fixed points is con
stant. Each of these fixed
points is called a focus
(plural foci).
Figure 44 shows an el
lipse for which the sum
of the distances of the
point P from the foci F
and F' is equal to the
sum of the distances of
any other point in the bounding line from F and F'. AB is
called the major axis and CD the minor axis.
Fig. 41
MENSURATION OF AREAS 127
To Construct the Ellipse. Place the given diameters AB
and CD at right angles to each other at their centres E.
From D with radius AE cut the major axis at F and F'. This
gives the foci. Place pins at F and F' and also at D. Place
a string around the three pins forming a triangle of string
FDF'. Take out the pin at D and, substituting a pencil,
trace as in Figure 44. If we represent the major axis by 2a
and the minor axis by 26 the circumference of the ellipse is
given by. the formula:
Circumference = tt (a \b) (approximately) .
The area of the ellipse is given by the formula Area = ■*■ ab.
100. Regular Polygons. A regular polygon is a figure
having all its sides equal and all its angles equal.
Fig. 45
ABCDE in Figure 45 is a regular pentagon (5 sides).
If we bisect the angles A and E, the point of intersection
of the bisectors will give us the centre of the inscribed
circle of the pentagon, and hence a point equidistant from
all the sides. This perpendicular from the centre on the side
is called the apothem. The area of the triangle AOE = \ar,
.*. area of pentagon = 5 X \ ar.
Generally, then, the area of the polygon with n sides is
\nar.
128
MATHEMATICS FOR TECHNICAL SCHOOLS
Since it is necessary to employ Trigonometry to find the
apothem, the following table is given:
Number of
Sides.
When Side is a
Multiply a by
When Area is a 2
Multiply a by
3
•433013
1519671
4
1
1
5
1720477
•762387
6
2598076
•620403
7
3633912
•524581
8
4828427
•455090
9
6181823
•402200
10
7 • 694209
•360511
11
9365640
•326762
12
11196150
• 298858
15
17642360
• 238079
18
25520770
• 197949
20
31567876
• 177980
Explanation of table. The first column gives the number*
of sides, the second gives the area when the side is known,
the third gives the side when the area is known.
Thus, if the side of a fivesided regular polygon is 6 in.
then the area is obtained by multiplying 6 2 by 1720477;
also if the area of a tensided regular polygon (decagon) is
256 sq. in. the length of a side is obtained by multiplying
•v/256 by 360511.
Example: — The side of a twelvesided regular polygon is 7".
Find the area.
MENSURATION OF AREAS
129
From the second column of the table —
Area = 7X7XlM96150 = 54861135 sq. in.
Example: — The area of a ninesided regular figure is 726
sq. ft. Find the length of a side.
From the third column the length of the side = V726 X • 402200
= 108353'.
101. Irregular Figures. Simpson's Rule for Finding Area.
The area of an irregular figure may be accurately determined
by the use of a planimeter, a description of which is given on
page 131. When great accuracy is not required, a sufficiently ac
curate measurement may be made by the use of Simpson's Rule.
Fig. 46
Figure 46 represents an irregular figure
divided into eight equal
parts. The perpendiculars
to this base line, d lf d 2 , d 3 ,
a 4 , a 5 , a 6 , a 7 , a 8 , a 9 , are
called ordinates, and since
there is an even number of
divisions there will be an
odd number of ordinates.
The rule applies only when
there is an odd number of
ordinates.
Consider Figure 47 consist
ing of the first two sections
of Figure 46.
HM is drawn through E
parallel to BC and such that BK = KN = NC.
The base line is
o
136 MATHEMATICS FOR TECHNICAL SCHOOLS
The area of ABCD = area of A BKH +area of HKNM +
area of MNCD approximately.
Area of AB KH = $s(d l +dJ
Area of HKNM=%sd 2
Area of M N C D = ±s(d 2 +d 3 )
Area of ABCD = \ s(d x +±d 2 +d 3 )
Similarly the area of the third and fourth sections
Similarly the area of the fifth and sixth sections
= i*(d 5 +4d 6 +d 7 )
Similarly the area of the seventh and eighth sections
= £s(d 7 +4d 8 +d 9 )
Adding these we get the total area
 s {(d 1 +d 9 )+W 2 +d i +d a +d 8 )+2(d a +d & +d 7 )}
= i*C4+4JS+2C)
Where A = sum of first and last ordinates.
B = sum of the even ordinates.
C = sum of the odd ordinates, omitting the first and
last.
s = common interval.
102. The Planimeter. The name of the instrument comes
from "planus" meaning flat, and "meter" meaning measure.
As the principle of recording area is the same in both of the
types shown, Figures 48, 49, we will confine our description to
the compensating planimeter. Its use consists in tracing the
contour of the figure to be measured with the tracer / as
shown. When doing so the wheel M is made to revolve, and
it is by the extent of these revolutions that the area of the
traced figure is ascertained. The various parts of the planimeter
are so dimensioned as to bring about one complete revolution
of the wheel when an area of 10 sq. in. has been traversed.
MENSURATION OF AREAS
131
Attached to the wheel is a white drum, divided into 100
parts, one of which indicates an area of • 10 sq. in. By means
r Hi. HJ
~A.MSI.KR PlaNIMETEH
Fig. 49— Otts Compensating Planimeter
of the vernier (see page 173) the single parts of the drum can
be read to tenths, giving an area of 01 sq. in. To keep a
record of the number of revolutions of the wheel a counting
disc Z is attached to it by means of a wormgear. Each mark
on the disc corresponds to one revolution of the wheel, there
fore 10 revolutions of the wheel corresponds to one revolution
of the disc.
132 MATHEMATICS FOR TECHNICAL SCHOOLS
Applying this to the reading in Figure 50 we have— the last
number on the disc is 3, .'. 30 sq.
in.; the last number on the drum is
5, .'. 5 sq. in.; the last division
2
^ between the 5 and 6 is 8, .". 8 sq.
^///////7?T^rf////////////////////Z. in.; the 4th division of the vernier
FlG * m is opposite a division on the drum,
.". 04 sq. in. Total reading 3584 sq. in.
Exercises LXIII.
1. Construct an ellipse in your laboratory book as suggested.
Write a note as to why your construction fulfils the require
ments. Find the area by counting the squares and check by
formula for area.
2. Construct an ellipse on cardboard having major and
minor axes 4" and 2" respectively, and also a rectangle having
length 4" and breadth 2". Cut out both, weigh, and find
the value of — ' — ^ — ^ — — . Do the same with different
wt. of ellipse
materials and find the average of your results.
3. A plot of ground in the form of an ellipse has major
and minor axes, 200' and 150' respectively. Draw to scale
in your laboratory book and find the perimeter and area.
4. An elliptic manhole door has major and minor axes of
3' and 2' respectively. It is made of cast iron \" thick.
Find weight if 1 cu. in. weighs 26 lb.
5. At what distance from the end of the major axis should
the hole for the centre of revolution be drilled in an elliptic
gear whose axes are \\" and 2"? (Elliptic gears will mesh
when revolving about their foci).
6. The area of an elliptic sheet of zinc is 88 sq. in. If its
minor axis is 4", find its major axis.
7. The head of a hexagonal bolt is \" to the side; find the
area of the head.
8. A square is 4" to the side. An octagon is formed by
cutting off the corners of the square. Find the side of the
octagon and hence its area. Find the area by subtracting
the areas of the four corners from the square and compare
with previous result.
MENSURATION OF AREAS
133
9. Ten hurdles, each 4' long, are placed to form a regular
decagon. Find the area enclosed.
10. A steel plate in the form of a regular pentagon measures
If " on each side and is \" thick. Find its weight, if a cu. in.
of steel weighs 283 lb.
11. The area of a regular hexagon is 284112 sq. in. Find
a side of the hexagon.
12. Regular polygons of 6 sides are inscribed in and circum
scribed about a circle of radius 1\ Find the difference of
their areas.
13. Construct a semicircle 4" in diameter in your laboratory
book. Find its area by Simpson's rule. Check by means of
formula for the area of a circle and thus calculate the per
centage error in Simpson's rule.
14. Construct an ellipse with major and minor axes 4" and
2" respectively. Proceed as in the preceding question.
15. Make a drawing of Figure
51 in your laboratory book. Com
mon interval f". If the scale
be £%" to the foot, find the area
in square ft. by Simpson's rule.
Check by planimeter and estimate
percentage error. Note — areas of
similar figures are to one another
as the squares on corresponding
sides.
16. The ordinates of a curved
piece of sheet lead in inches are
20, 30, 299, 295, 284, 257, 142. The common distance
between them is 365"; find the area.
17. The halfordinates of a transverse section of a vessel
are in feet 122, 122, 121, 118, 112, 10, 73 respectively.
The common interval is 18"; find the area.
w
f
p
1<
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p
WS L
^
"A
J J * * p. iy
/'V/' Jf .o
Fk
3. 51
CHAPTER XI.
RATIO AND PROPORTION.
103. Ratio. We are constantly comparing weights, dis
tances, sizes, etc. If one piece of metal weighs 50 lb. and
another 10 lb., we say that the first is five times as heavy as
the second, or that the second is onefifth as heavy as the
first. If one board is 8 ft. long and another 2 ft. long, we
say that the first is four times as long as the second, or that
the second is onefourth the length of the first.
This Relation between Two Quantities of the same Kind is
called Ratio.
Note. — In the above definition of ratio it is important to
notice "of the same kind." It would clearly be absurd to
compare bushels and feet.
A ratio may be written in two different ways. For example,
the ratio of the diameters of two wheels which are 10 in.
and 16 in. in diameter can be written as a fraction ff. Again,
since a fraction indicates division, i.e., 10r16, the line in
the division sign is sometimes left out and the ratio is written
10 : 16. In either case the ratio is read "as ten is to sixteen."
Since a ratio may be expressed as a fraction it may be
reduced to lower terms without changing its value. For
example, if one casting weigh 600 lb. and another 150 lb., the
ratio of the weight of the first to the weight of the second
id 6004
1S T5~ff — T
Example 1:
The diameter of the cylinder on an engine is 18" and the
diameter of the piston rod is 3". What is the ratio of the
cylinder diameter to the piston rod diameter?
Diameter of cylinder 1 8 _
Diameter of piston rod ~ s ~ 1
134
RATIO AND PROPORTION 135
Example 2:
A concrete mixture is made of cement, sand, and gravel in
the ratio of 1 : 2\ : 5. If 25 bags of cement be used (1 bag
= 1 cu. ft.), how many cu. ft. of sand and gravel will be
required?
25 (cu. ft. cement) X2§ = 62£ cu. ft. sand.
25 (cu. ft. cement) X 5 =125 cu. ft. gravel.
104. Proportion. When two Ratios are Equal, the Four
Terms are said to be in Proportion.
The two ratios 3 : 9 and 12 : 36 are evidently equal, since we
can reduce 12 : 36 to 3 : 9.
When written 3 : 9 = 12 : 36, these numbers form a proportion.
Further we observe in the above proportion that the pro
duct of 3 and 36, i.e., the first and last, is equal to the product
of 9 and 12, i.e., the second and third.
The first and last are called the extremes, and the second
and third are called the means. We, therefore, have : — The
product of the means is equal to the product of the extremes.
This relation may be expressed generally.
If a, b, c and d represent the four terms of any proportion
then:
,, extremes N
a : b = c : d
^means^
Then in accordance with the above ad = bc.
Application of this principle to some practical problems.
Example 1:
If it requires 60 men to turn out 200 shells in a day, how
many men will be required to turn out 360 shells in a day?
If x be the required number of men, then 60 : x = 200 : 360
or 200z = 60X360
60X360
X ~ 200 ~ 108 '
136
MATHEMATICS FOR TECHNICAL SCHOOLS
Example 2:
If the diameter of a pulley is 40", and it makes 120R.P.M.,
what is the R.P.M. of a second pulley belted to the first if
its diameter is 16"?
Note. — When two pulleys are belted together, the larger
of the two is the one that makes the least R.P.M. The pro
portion formed from their diameters and revolutions is,
therefore, called an inverse proportion.
Letx = R.P.M. of the second pulley, then 40: 16 = .r:120
or 16x = 40X120
40X120
or s =
16
= 300
After some practice in proportion we might write this
directly— R.P.M. of 16 in. pulley = f£ of 120 = 300.
105. Proportion in Similar Triangles :
In triangle ABC and DEF, BC and EF are \\" and 3"
respectively, also Z B = Z E and ZC= Z.F.
The triangles ABC and DEF are, therefore, equiangular
and are called similar triangles. If we compare corresponding
sides with dividers we observe that DE = 2AB and DF = 2AC.
RATIO AND PROPORTION
137
This experiment would suggest that, when triangles are
equiangular, their corresponding sides are proportional.
BC = AB = AC
' EF DE DF'
Make drawings in your laboratory book to verify the
above.
Observe this principle in the following Example :
Fig. 54
The top of a telegraph pole, Figure 54, is sighted across a
5' pole placed 100' from the foot of the telegraph pole, the
observer sighting from the ground at a distance of 15' from
the foot of the 5' pole. Find the height of the telegraph pole.
The triangles ABE and CDE are equiangular and therefore
similar.
CD = DE
'*■ AB~BE
CD 115
5 " 15
or, CD=^X^=38*'.
Exercises LXIV.
1. A room is 16' by 12'. What is the ratio of the length
to the breadth?
2. Two gearwheels have 100 teeth and 40 teeth respectively.
What is the ratio of the number of teeth?
3. Detroit has a population of 1,000,000 and Toronto
600,000. What is the ratio of the population of Toronto to
that of Detroit?
or,
138 MATHEMATICS FOR TECHNICAL SCHOOLS
4. A man rode 280 miles partly by rail and partly by boat.
What distance did he travel by each, if the ratio is as 3 to 2?
5. A locomotive has a heating surface of 1340 sq. ft. and
a grate area of 24 sq. ft. What is the ratio of the heating
surface to the grate area?
6. The steam pressure in a locomotive is 196 lb. and the
mean effective pressure in the cylinders is found to be 80 lb.
What is the ratio of the mean effective pressure to the boiler
pressure?
7. Two pulleys connected together have diameters of 36"
and 22". If the first makes 120 R.P.M., what is the R.P.M.
of the second?
8. The ratio of two gears connected together is as 5 to 3.
The first makes 105 R.P.M. , how many does the second gear
make?
9. If 15 tons of coal cost $120, what will 18 tons cost?
10. If the freight charges on a shipment be $40.50 for
216 miles, what should it be for 300 miles?
11. A pump discharges 20 gal. per min., and fills a tank
in 24 hrs. How long would it take to fill the tank with a
pump discharging 42 gal. per min.?
12. A machinist gets $7.50 a day and a helper $5.00 a day.
How much would the helper receive when the machinist gets
$80.00, providing they both work the same number of days?
13. If a yardstick held upright casts a shadow 3' 9" long,
how long a shadow would be cast at the same time by a
chimney 66' 8" high? Check by drawing to scale.
14. What is the height of a signal pole whose shadow is
12', when a 10' pole at the same time casts a shadow of 2'
4"? Check by drawing to scale.
15. The length of shadow of a telegraph pole to the first
crossarm is 24'. A 6' pole at the same time casts a shadow
of 2' 8". What is the height of the first crossarm? Check
by drawing to scale.
16. In a mixture of copper, lead and tin there are 4 parts
copper, 3 parts lead and 1 part tin. How many lb. of each
would there be in 248 lb. of the mixture?
17. A solder is made of 5 parts zinc, 2 parts tin, and 1 part
lead. How many parts of each metal in 96 lb. of the
mixture ?
RATIO AND PROPORTION 139
18. A man 6' high, standing 8' from a lamppost, observes
his shadow to be 6' in length. Find the height of a boy who
easts a shadow of 4' when he stands 9' from the lamppost.
19. What will be the diameter of a gear that is to make
60 R.P.M., if it is to mesh with a gear 24" in diameter, which
makes 72 R.P.M.?
20. What is the percent, grade of a road bed that rises
12' in a horizontal distance of 40'?
21. The roof of a house rises 1 # in a run of 2'. How far
will it rise in a run of 20'?
22. A road bed rises 2\' in 200', what is the percent, grade?
In how many feet will it rise 1'?
23. A cable railway up the side of a mountain has at places
on the route a grade of 20%. What rise does this represent
for every mile of horizontal distance?
CHAPTER XII.
SIMULTANEOUS EQUATIONS.
Formulas — (continued).
106. In the Discussion of the Simple Equation, we learned
that it contained only one unknown quantity and that,
therefore, the value of the unknown could be definitely de
termined.
Thus, if 3.r+4 = 16
3* = 12
x = 4.
If, however, a single equation contains two unknown
quantities, we cannot find a definite value for either of them,
and are restricted to finding the value of one in terms of the
other.
Thus, in the equation x\2y = 13, if we transpose 2y we
have £ = 13 — 2y. The value of x in this equation will depend
upon the values assigned to y.
If, y = l, x = 13 2 = 11.
y = 2, x = 13 4= 9.
2/ = 5, a; = 1310= 3.
Similarly, if we give values to x we may obtain corresponding
values for y. It is evident, then, that we cannot find definite
values for x and y but have merely a statement of relation
between them.
If, however, another relation between x and y be obtained
from the same problem, we can, from the two equations,
determine the definite values of x and y. If we also find
that 3z+2/ = 14, then transposing and dividing by 3 we have
Uy
T =
x 3 .
140
SIMULTANEOUS EQUATIONS 141
If, as above, y = l, x =
141
1 3
3
O 14 ~ 2 A
y = 2, x=^—=4:.
145
y = o, x = —^—=3.
Comparing the two sets of values for x and y we observe
that there is only one pair of values that will satisfy both equa
tions, namely, x = S and y = 5. These values of x and y for
the two equations are called simultaneous values and the
equations are known as simultaneous equations.
107. In the following Problem we will illustrate Two Methods
of Solution. One alloy contains 4% copper and another
10% copper. How many pounds of each should be used to
make a 100 pound mixture containing 6% copper?
Solution by Substitution:
Let # = No. of lb. from the first alloy.
And ?/ = No. of lb. from the second alloy.
Then x+y = 100. (1)
4x 1 Oy 6
also, Tnrf+TQA  = 7oo~ xl00 > which reduces to 4x + 10?/ = 600
or, 2x + 5?/ = 300. (2)
From (1) a: =100 y
Substituting in (2)
2 (100i/) +% = 300
2002?/ +5y = 300
giving y = 33f.
Substituting in (1) a: = 66f.
Therefore we must use 66f lb. from the first alloy and
33J lb. from the second.
142 MATHEMATICS FOR TECHNICAL SCHOOLS
Solution by Elimination :
x+ y = 100 (1)
2z+5?/ = 300 (2)
(l)X2 = 2x+2?/ = 200
(2) = 2x+5y = Z00
Subtracting 3y =  100
2/ = 33i
and x = 100?/ = 66§.
In simultaneous equations it frequently happens that one
of the unknowns can be readily expressed in terms of the
other, and the solution obtained by means of the simple
equation. The problem given above affords an example of this.
It has been observed that, if the equation contains only one
unknown, the value of this unknown may be found from one
equation; also that, if the equation contains two unknowns,
two equations are necessary to find the values of the unknowns.
This may be extended to three or more, and we say,
generally, that we must have as many distinct equations
as there are unknowns to be found.
Exercises LXV.
Solve the following equations:
1. 3*+4y = 10. * ,* «
4x + y = 9. 7  5 + 2~ 5 '
2. 4x+7?/ = 29. xy = ±.
*+3y=U. *a.»ijB
3. 8s y = U. 8^3
x+8y = 53. *_y = A
4. 2x+oy = 25. 4 5
3x y = 0. 2 _
5. 3x5y = 6. 9 ' x +y = 1 '
2x X *
6. ^+ y = 16.
10. f!=0.
J "4~"" 3x£y = 17.
■+IM
SIMULTANEOUS EQUATIONS 143
Exercises LXVI.
1. If scrapiron contains 4% silicon and we have a pigiron
containing 9% silicon, how many pounds of each must be
used to make a ton of mixture containing 6% silicon?
2. The law of a machine is given by R = aE\b and it is
found that when fl=100, £ = 25, and when R = 250, £ = 60,
find a and b.
3. One mixture for casting contains 20% copper and
another mixture, for the same purpose, contains 8% copper.
How many pounds of each should be taken to make 200
pounds of a mixture containing 12% copper?
4. The distance between the centres of two parallel shafts
is 8". It is required to connect these shafts by a pair of
gears so that one shaft will turn twice as fast as the other.
Calculate the diameters of the gears.
5. In a pulleyblock lifting tackle, a force of 15 lb. will
lift a load of 100 lb., and a force of 35 lb. will lift a load of
300 lb. If the force (P lb.) and the load (W lb.) are related
by an equation of the form P = mW\k find the values of m
and k, and hence the law of the machine.
6. If E represents the fixed expenses of a manufacturing
company, V the variable expenses for each machine manu
factured, .V the number of machines per year, and C the
total cost of operating, then C= E\VN. In 1918 the com
pany built 600 machines at a total cost of $18,000, and in
1919, 800 machines at a total cost of $22,000. Calculate
E and V.
7. The law of a machine is given by E = aR\b and it is
found that when ii = 10, £ = 546 and when R = 100, E =
96; find a and b.
8. The total cost C of a ship per hour is given by C = a\bs z
where s is the speed in knots. When s is 10, C is found to be
$26.00 and when s = 15, C is found to be $36.50. Find a and
b and express the relation between C and s.
9. At an election there were two candidates and 3478
votes were cast. The successful candidate had a majority
of 436. How many votes were cast for each?
10. The moment of a force is the tendency of that force
to produce rotation of a body, and is measured by the product
of the force (in pounds) and the perpendicular distance (in
feet) from the axis to the line of the applied force. For
144 MATHEMATICS FOR TECHNICAL SCHOOLS
equilibrium in the case of parallel forces, the algebraic sum
of the forces is and the algebraic sum of the moments is 0.
A uniform plank 20' long, weight 90 lb., rests on supports
at its ends. A load of 500 lb. rests 8' from one end. Find
the reactions of the supports.
11. A uniform beam 16' long weighs 300 lb. It is supported
at one end and at a point 4' from the other end. Calculate
the reactions of the supports.
12. A uniform beam, 12' long, is supported at each end and
carries a distributed load, including its own weight, of \ ton
per foot run. A concentrated load of 1 ton rests 5' from
one end and another of 3 tons, 4' from the other end. Calculate
the reactions of the supports.
13. If ,f, =34 and  = 147 find E and r.
240+r r
14. If Vi=V {l + Bt) and 1^ = 124 when * = 215 and
^ = 173 when t = 7o0, find V and B.
15. The receipts of a railway company are divided as
follows: — 40% for cost of operating; 10% for the reserve fund;
a 6% dividend on the preferred stock which is j of the capital;
and the remainder, $630,000, as dividend on the common
stock, being at the rate of 4% per annum. Find the capital
and receipts.
Exercises LXVII.
Formulas — (continued) .
1. The time taken by a pendulum for a complete oscillation
is given by 1 = 2tt\  , where t is the time in seconds, I the
length in ft. and g the acceleration due to gravity in ft. per
sec.
Solve for I and g:
If / = ii, = 322 find t.
If t = 2, g = 322 find I
If * = l57, 1 = 2 find g.
2. The resultant of two forces at right angles is given by
R= y/P 2 +Q i , where P and Q are the forces at right angles and
R the resultant.
Solve for P and Q:
If P = 8, Q = 5, find R.
If R = 17, Q = 8, find P.
SIMULTANEOUS EQUATIONS 145
3. The velocity of a body at the end of a specified time is
given by v = u\at, where v is the final velocity in ft. per sec,
u the initial velocity in ft. per sec, a the acceleration in ft.
per sec per sec., t the time in seconds.
Solve for v, u, a, t:
Iiu = 12, a = lo,t= 6, find v.
If v = 750, a = 30, t = 18, find u.
If t' = 109, w = 45, < = 16, find a.
Iff 215, u = 7o, a = 10, find*.
4. The space traversed by a body is given by s= ut+%at 2 ,
where s, is the space in ft., u the initial velocity in ft. per
sec, t the time in sec, a the acceleration in ft. per sec.
per sec.
Solve for u and a:
Ifu = 20, a = 322, t= 8, find 5.
If 5 = 300, a = 16, t= 5, find w.
If s = 750, u = 25, t = 10, find a.
5. The thickness of plate required in a boiler is given by
'pd
t = ^j , where t is the thickness in in., p the pressure in lb. per
sq. in., d the diameter of the boiler in in., / the tensile stress in
lb. per sq. in.,
e the efficiency of the joint.
Solve for p,
d, f, and e
If p = 160,
d = 8ft.,
/ = 20000, e = 7,
find *.
If t=o,
d= 90,
/= 16000, e=6,
find p.
Tf / — 3
ii t — 8 ,
p = 150,
/= 18000, e=7,
find d.
If / = *
A* «< — 16 >
p = 140,
e =75, d = 72,
find/.
If < = A
Al t — 16 ,
p = 120,
d = 48, / = 5tons
find e.
6. The Kinetic energy of a falling body in footpounds is
wv ^
given by K = ^— , where K is the energy, w the weight of the
body in lb., i; the velocity in ft. per sec, g the acceleration due
to gravity.
Solve for iv, v, g:
Ifw = 1000, v = 44, = 322, find K.
li K=\12, w = 5000, y = 322, find v.
IfX=124, t> = 35, = 322, find w.
146 MATHEMATICS FOR TECHNICAL SCHOOLS
7. The effort of friction (measured in lb.) in diminishing the
load lifted is given by E=PV—W, where E is the effort of
friction, P the effort in lb., W the weight in lb., and V the
, ., ,. motion of effort
motion of weight'
Solve for P, V, W:
IfP = 24, F=16, W = 25,
find E.
If E= 52, F = 16, P =42,
find W
l(E= 82, F = 16, W = Z0,
find P.
If £=104, P= 9, JF = 40,
find V.
8. The magnetic lines of force (Flux) is given by Q % = — s — ,
K
where Q is the total flux, N the number of turns of wire in the
coil, R the reluctance of the magnetic circuit, / the current in
amperes.
Solve for N, I, R:
UN = 200, 7 = 5, fl=0002, find Q.
If Q = 30000, iV = 500, 7=15, find R.
If Q = 100000, N = 50, R= 00005, find I.
9. For a single riveted lapjoint, the efficiency in tension is
P — d
given by K t = — =— , where i£, is the efficiency in tension, P
the pitch in in., d the diameter in in. of the rivet.
Solve for P and d:
If P=H in., d= f in., find K t .
If #,= 5, d= A in., find P.
If #,= 6, P=lf in., find d.
10. The relation between a Centigrade and a Fahrenheit
scale is given by C = f(P — 32), where C represents the Centi
grade and F the Fahrenheit reading.
Solve for F:
If F = 63°, find C.
If C = 72°, find P.
If C=4°, find P.
11. The counter electromotiveforce (E.M.F.) of a motor is
given by E= E C + IR, where E is the impressed E.M.F. , E c
the counter E.M.F., /the current in amperes, R the resistance
in the armature.
R =
•08,
find E.
R =
•05,
find E,
I =
25,
find R
R =
•15,
find /.
SIMULTANEOUS EQUATIONS 147
Solve for E c> I, R:
If E c =108, I =40,
If E =220, I =60,
HE =110, E c = 1085,
If E =110, £ c = 109,
12. The approximate length of an open belt connecting two
wheels is given by L = %\{R\r)\2d, where L is the length,
R and r the radii of the large and small wheels respectively,
d the distance between the centres.
Solve for R, r, d:
If #=18 in., r = 10in., d = 40in., find L.
If L= 15 ft., # = 16 in., r = 12in., find d.
13. The approximate length of a crossed belt connecting two
wheels is given by i = 3f (R\r) \2d, where L, R, r, and d
have values as in question 12.
Solve for R, r, d:
If #=18 in., r=12in., d = 6ft., find L.
IfZ = 12ft., # = 16in., r=8in., find d.
14. The horsepower transmitted by belts is given by H.P.
(T — T )V
= qq nnn — » where 7\ is the tension on the tight side of the
belt in lb., T 2 the tension on the slack side of the belt in lb.
V the velocity in ft. per minute of the driver.
Solve for T,T 2 , V:
If T l =120, r 2 = 50, F = 31416, find H.P.
UH. P. = 81, V =2500, findr,r 2 .
15. The width of a single belt required to transmit a given
horsepower at a given speed of the belt is expressed by
W = — ' — , where W is the width in in., H.P. the horse
power, S the speed in ft. per min.
Solveforff.P. and S:
II H.P. = 100, S =3000, find IF.
IfS =3200, IT = 6 in., findtf.P.
Iftf.P.=40, IF = 5 in., find S.
16. For a double belt the formula in question 15 becomes
w = HJ\X?ti000
SX100 '
148 MATHEMATICS FOR TECHNICAL SCHOOLS
Solve for H. P. and S:
If #.P. = 100,
S =3000,
find W.
If 5 =3200,
W = 6in.,
find H.P.
If //.P. = 50,
W = 5 in.,
find S.
17. The length of belting in a closely rolled coil is given by
i= 1309 N (D+d), where L is the length in ft., D the diameter
of the roll in in., d the diameter of the eye in in., N the number
of turns in the coil.
Solve foriV, D, d:
IiN= 15, D = 16h d = 5, find L.
If L= 80, D = 14, d = 3, find N.
If Z = 200, D = 44, iV = 30, find d.
18. For a single riveted lapjoint the efficiency in shear is
given by K s = p ' , where l£ s is the efficiency in shear, a
the crosssection area of the rivet in sq. in., P the pitch of the
rivet in in., T the thickness of the plate in in., S s the strength
of rivet steel in shear (lb. per sq. in.), S t the strength of
plate in tension (lb. per sq. in.).
Solve for a, S s , P, T, S t : ,
If a= 7854, S s = 30000, P=Hin., 2T = f in.,
^ = 40000, find X s .
IfX 4 = l5, & = 32000, P = lfin., I 7 = 5 in.,
«S, = 35000, find a.
IfX, = l75, a =5, P = l3 in., T = \ in.,
'«, = 38000, find S,.
19. The horsepower of a boiler is given by B.H.P.=
Q . K N/QA g 7 > where B.H.P. is the boiler horsepower, W the
number of pounds of water evaporated per hour, H the total
heat of steam above 32° F., t the temperature of the feed
water.
Solve for W, H,t:
If W =20000,
ff=1180,
t = 100°,
find B.H.P
If B.H.P. =600,
77 = 1175,
t = 120°,
find W.
If B.H.P. = 650,
H = 1200,
W = 20000,
find/.
SIMULTANEOUS EQUATIONS 149
20. The quality of steam (%) as determined by the throttling
calorimeter is given by s = 100{ *L s }, where x
is the moisture in steam, H the total heat of steam at
main pressure, h the total heat of saturated steam at pressure
in calorimeter, T e the temperature of saturated steam at
pressure in the calorimeter, T s the observed temperature in
the calorimeter, C p the specific heat of superheated steam at
constant pressure, L the latent heat of steam at main pressure.
Solve for H, h, C p , T s , T e , L:
Ifff=1180, A = 1150, r s = 220, T e = 215, C p =48,
Z = 920, finds.
lix = ^{2%), A = 1160, 2^ = 225, r c = 218,
C=48, £ = 930, find H.
CHAPTER XIII.
GRAPHS.
108. If we wish to fix the position of a point P on the page
of this book, one way would be to find its perpendicular distance
from the left of the page and also its perpendicular distance
from the bottom of the page. If these distances were 3 in.
and 4 in. respectively, then the point P would be definitely
fixed with respect to the plane of the paper.
Consider a sheet of paper ruled as in Figure 55. If we know
that a point P is 6 divisions to the right of Y and 4 divisions
above OX, we can, at once, locate the position of the point
by counting 6 divisions along OX and then counting 4 divisions
vertically to the point P. This method of fixing the point is
called plotting the point. The lines OX and OY are called
Axes of Reference, the point of intersection is called the
Origin, and the distances 6 and 4, which locate the point, are
called Coordinates. We would now say that the coordinates
of P are 6 and 4, and would write it P (6, 4), the first number
always giving the distance along OX and the second the distance
along OY. OX is usually spoken of as the axis of X and
Y as the axis of Y. Distances along OX are called abscissae
and distances along Y are called ordinates. We see from the
above that any point can be plotted on the squared paper if
we know its distances from the axes Y and X.
109. Let us use this for a Practical Purpose. A sewer runs
across a rectangular lot and it is necessary to know its exact
location in case of trouble later.
150
GRAPHS
161
A to cixv
Fig. 55
152 MATHEMATICS FOR TECHNICAL SCHOOLS
_j uu ^4 1 ,  i , £ :_i __ i ,
1!!==!=!==!=!===!; ' i ■ ■' :==: :' \^
Fig. 56
GRAPHS 153
Measurements are taken according to the following plan:
Distance from Y
5
10
15
25
35
45
55
65
75
Distance from OX
5
10
20
30
40
50
60
70
A graphical representation of the position of the sewer is
shown in Figure 56. Each small division of the squared
paper represents 1 ft. The first point A has for its coordinates
(5, 0), the second point 5(10, 5), the third point C(15, 10)
and so on. If we join these points we have a graph of the
position of the sewer. At some subsequent date it is necessary
to make an excavation for the footings of a building on this
lot, and the contractor wishes to know if a particular footing
will come too near the sewer. He takes measurements and
finds that the distance from the side corresponding to O Y is
30 ft., and the distance from the side corresponding to OX
is 25 ft. While these distances are not actually recorded in
the data previously taken, yet by going out 30 ft. (30 spaces)
from OY and up 25 ft. (25 spaces) from OX, he would find
that he is directly over the sewer. This illustration brings
out one of the most important functions of a graph: It
gives results for data not actually recorded at the outset.
110. It usually happens in practice that we require to make
a record of two corresponding sets of measurements, in which
the unit of measurement in one is entirely different from the
unit in the other.
The following will illustrate: .
The observations below were taken of the loads on a
lighting plant from 3 P.M. to 12 P.M. at intervals of one
hour.
Time in hours. . . .
3
4
5
6
7
8
9
10
11
45
12
Load in Kilowatts.
50
60
76
120
140
150
142
100
30
164
MATHEMATICS FOR TECHNICAL SCHOOLS
S * 3 «
Fig. 57
GRAPHS
155
Figure 57 shows how the relation of time and load
may be represented graphically. , Along the axis of X we let
each main division represent one hour, while along the axis
of Y we let each main division represent 20 kilowatts. The
letters, A, B, C, D, E, F, G, H, I, J, represent the locations
of the observations. By drawing a curve through these
points, we have a graph which will show at a glance the varia
tions in load. We see, further, that it will give us the probable
load for times in between those recorded in the data above.
For example, if we wished to know the load at 8.30 P.M. we
would take a point halfway between 8 and 9 on the axis of
X and draw a perpendicular to it, represented by the heavy
line in the figure.
From the point where this meets the curve, draw a line
perpendicular to the axis of Y as represented, and we have
148 kilowatts as the probable load at 8.30 P.M.
111. Frequently a Relation between Measurements is
expressed by an Algebraic Equation. Suppose we wished to
find the Fahrenheit reading corresponding to a Centigrade
reading in degrees. Since 180 degrees Fahrenheit, measuring
the range from 32° to 212°, are equal to 100 degrees Centigrade,
measuring the range from 0° to 100° we have :
100° Centigrade = 180° Fahrenheit.
a° Centigrade = £# a° Fahrenheit =  a° Fahrenheit.
If 6° represent the Fahrenheit reading corresponding to a°
Centigrade, then the relation is given by the equation b =fa+32.
By giving different values to a in the equation we can obtain
the corresponding values of b. These may be tabulated as
follows :
Values of a
32
10
50
20
30
40
60
80
100
Corresponding values of b
68
86
104
140
176
212
156
MATHEMATICS FOR TECHNICAL SCHOOLS
Figure 58 is a graphical representation of this algebraical
relation. Along the axis of Y we have represented the values
of a, while along the axis of X we have the corresponding
values of b. Corresponding values other than those recorded
may be read off from the graph. Thus 77° Fahrenheit equals
25° Centigrade as represented in the drawing.
Another value of graphs is here illustrated, that is, they
act as checks on computations.
112. It is often important to represent more than one Set of
Relations on the same Sheet.
The following are the results obtained with a wheel and
axle mounted on ordinary plain bearings. W represents the
load lifted in pounds, P the effort applied in pounds, F the
friction measured in pounds, E the efficiency percent.
w
P
F
E
•8
160
5
430
360
582
10
714
428
701
15
991
482
757
20
1281
562
780
25
1563
626
800
30
1850
700
812
35
2150
800
814
40
2445
890
818
GRAPHS
157
Fig. 58
158 MATHEMATICS FOR TECHNICAL SCHOOLS
4+H
I i
■ '•" "i  — : — i — i — ; —
ZO 25 »<>
LOA.PVH *OUMP»
Fig. 59
In Figure 59, the lower line represents the relation between
the load and the effort. The middle line represents the
relation between the load and the friction. The top line
represents the relation between the load and the efficiency
percent.
Note.— In the drawing of a graph relating to machines it
often happens that the points are not absolutely on a straight
line. It is necessary, in such a case, to take the line which
lies most nearly along the path of the points.
GRAPHS
159
(<SNOi. OOSI a3Ao)dlHC J — «NO\SIAla IITHS z
Fig. 60
113. Sometimes the sole Purpose of a Graph is to picture in a
concise and striking way the relation between two measurements.
Figure 60 is a graph (from a Toronto daily paper) based on
the official weekly figures of losses sustained by the British
merchant fleet during the height of the submarine warfare.
160
MATHEMATICS TOR TECHNICAL SCHOOLS
Fig. 61
GRAPHS
161
Fig. 62
162 MATHEMATICS FOR TECHNICAL SCHOOLS
114. In Business Transactions frequent use is made of the
Pictograph. This kind of representation takes a variety of
forms — varying sized men may represent populations, varying
sized bales of cotton may represent the export of cotton, and
so on.
Figure 61, called the bar pictograph, is of frequent use.
Example :
The net income of a certain railway company for a recent
year was divided as follows:
Sinking Fund Requirements, $2,000,000.
Dividend on Preferred Stock, $5,000,000.
Dividend on Common Stock, $18,000,000.
Additions and Improvements, $2,400,000.
Surplus to Profit and Loss, $6,200,000.
In Figure 61, the above amounts are represented by a series of
parallel bars, each main division on the vertical line representing
$2,000,000. When the division of the income was presented
in this form, the directors saw at a glance the relative division
of the returns from the road.
This form of pictograph is also extensively used to represent
a decline or growth in business.
Example: — The graph on page 161 is from a report re
Toronto's gross funded debt 19101919.
It illustrates clearly the rapidity of the growth in recent
years and its arrest in 1919.
When it is necessary to represent a percentage division, the
circular pictograph is of common use.
GRAPHS 163
Example. — The following figure is also from a report re
Toronto's debt for 1919:
MWJ .RfevEKlOE PRODUCVNG ceoSS TUMOEO DEBT
' 1 KokRevekuE PRooocinq f> iai,8i3,7&3
Fig. 63.
The object of this figure is to show the percentage of the
debt, which is due to each investment mentioned.
Exercises LXVIII.
Note. — Before attempting to draw a graph of any relation, it is
important to make a careful study of the squared paper at
our disposal for the work. The larger the graph the greater
accuracy and range of readings; we should, therefore, draw
our graph to cover, if possible, all the sheet. Further, we
should so divide the horizontal and vertical distances that as
many as possible of our readings may come exactly on the
lines of the paper.
164
MATHEMATICS FOR TECHNICAL SCHOOLS
1. The distances along a road from a certain point and the
height of the road above sealevel at these distances are shown
as follows:
Distance from start
ing point in miles.
1
2
3
4
5
6
7
8
Height above sea
level in feet
60
75
90
140
175
230
260
290
330
Represent the above relations by means of a graph. Esti
mate the probable height above sealevel 1\ miles from the
starting point.
2. The following table represents the output of an auto
mobile firm for the past ten years:
Year
1910
1911
1912
1913
1914
1915
1916
1917
1918
1919
Number. . . .
700
780
850
900
1000
950
960
1020
1050
1850
Represent the above relations graphically.
3. The following table gives the revolutions per minute of a
60 in. diameter locomotive driver and the corresponding
speed of the locomotive in miles per hour:
Revolutions per
min
60
90
100
150
200
250
275
300
Miles per hour. .
105
157
175
263
35
393
482
525
Represent the above graphically and find the revolutions
for a speed of 30 miles an hour.
4. The following observations of temperature were recorded
on July 25, 1919:
Hour of the day
4 A.M.
6 A.M.
8 A.M.
10 a.m.
12 NOON
2 P.M.
4 P.M.
6 P.M.
8 P.M.
10 P.M.
Temperature in
40
45
60
70
75
85
90
80
64
60
Draw a graphical representation of this variation in tem
perature.
5. If a cu. in. of steel weighs • 28 lb. construct a graph showing
relation between volumes and weights.
GRAPHS
165
6. If 1 inch = 254 centimetres, construct a graph showing
relation between the two systems of measurement.
7. The following is an extract from a table giving breaking
strength of steel, in pounds per sq. in., in relation to the per
centage of carbon in the steel:
.09
.18
.20
.31
.39
.50
.57
.71
.79
Breaking Strength
53000
64000
05000
77000
90000
97000
110000
124000
127000
Represent the above graphically and estimate the percentage
carbon for a breaking strength of 100,000 pounds per sq. in.
8. The prices charged by a manufacturing concern for a
certain motor of different horsepowers is given by the following
table :
H.P
1
2
3
4
5
74
' 2
10
15
Price
100
140
165
180
200
250
275
325
Represent graphically the relation between H.P. and price.
9. The quotations of a certain industrial stock at intervals
of a week, were, 48, 49, 52$, 53, 56, 58, 56$, 55$, 53.
Represent graphically the probable fluctuations in price.
10. The record of a patient's temperature for a certain time
at intervals of a halfhour, is 97, 975, 98, 98$, 99$, 101, 101$,
102, 101, 100. Represent the fluctuations graphically.
11. A tramcar is found to travel the distance y feet in x
seconds, the distance moved in different times being measured
and recorded as follows:
Distance in feet
(y)o
75
13
20
27
34
42
495
575
Time in seconds
(x)0
1
2
3
4
5
6
7
8
Represent this relation graphically.
12. A company finds that the buying expenses are 15% of
its gross income; office expenses 5%; management 10%; other
overhead 25%; selling expenses 30%; interest 10%; dividends
3%; incidentals 2%. Use the pictograph to represent the
division of the gross income.
166
MATHEMATICS FOR TECHNICAL SCHOOLS
13. The value of the exports and imports of the United
States for a given period is as follows:
Year
1830
1840
1850
1860
1870
1880
1890
1900
Value in
Millions. .
134
222
318
687
829
1504
1647
2100
Use the pictograph to represent this growth in commerce.
14. The following results were obtained by hanging a
series of weights on the free end of a spiral spring and thereby
stretching it:
Weight in lb.
1
•2
2
•4
3
•6
4
•8
5
6
7
8
9
10
Stretch in in.
10
12
14
16
18
20
Represent this relation graphically and indicate the probable
stretch for a load of 5 lb.
15. The following results were obtained as in the preceding
except that the stretch in inches is given by differences:
Stretch in in.
by differences
Plot the above in two parts — the first for loads up to 60 lb.,
the second for loads above. Compare the two graphs.
16. A car starting from rest is drawn by a varying force F
pounds, which, after t seconds, is as shown in the following
table:
t (seconds)
2
5
8
11
13
16
19
20
F (pounds)
1280
1270
1220
1110
905
800
720
670
660
If the frictional resistance is constant and equal to 500 lb.,
draw a graph of the above relation and indicate the force after
10 seconds.
17. The elasticity of a wire may be found by twisting. The
following readings were taken in experimenting with a steel
wire:
GRAPHS
167
Load in lb
1
2
4
5
6
Angle of twist in
degrees
6
12
24
29
26
Represent graphically and indicate the probable twist for
a load of 45 Tb.
18. The law of a machine is given by the relation P = • 08
FF+14. P being the force in pounds required to raise a
weight W. The following values of W are given: — 21, 3625,
662, 875, 10375, 120, 1525. Find the corresponding
values of P and plot the relation. Find the force necessary
to raise a weight of 310 lb. from your graph.
19. In a certain machine, P is the force in pounds required to
raise a weight W. The following corresponding values of P
and W were obtained experimentally:
p
28
37
48
55
65
73
80
95
104
1175
w
200
250
317
356
450
524
575
650
710
825
Draw the graph connecting P and W , and read the value of
P when FT = 70. Also determine the law of the machine, and
from it the weight that could be raised by a force of 45 pounds.
(P=aW+b).
20. The length of one degree on a parallel of latitude is given
for certain latitudes as follows:
Latitude. . .
10°
20°
30°
40°
50°
60°
70°
80°
90°
Length in
Miles.. . .
692
681
65
60
53 1
446
347
237
121
Draw a graph of the above and employ it to estimate the
length of a degree in latitudes 15°, 45°, 73°.
21. The following are the results obtained with a set of
rope pulleys:
5.5
12.2
17.1
2.5.0
31.0
37.5
44.8
50.8
62.0
76.0
Effort in lb
.94
3
5.5
7.1
10.2
12.2
14.5
17.2
19.7
25.5
30
3 76
6.5
9.8
11.3
15.8
17.8
20.5
24.0
28.0
40.0
43 9
45.8
55.8
60.2
61.3
63.5
64.6
65.2
64.5
60.8
63.4
168
MATHEMATICS FOR TECHNICAL SCHOOLS
On the same sheet of paper draw graphs of the relation
between load and effort, load and friction, load and efficiency.
From your graph estimate the effort necessary to lift a load of
40 lb., also the friction and efficiency for this load.
22. From a series of tests on an oil engine the following
values of the weight of oil used per hour (W) and the Brake
Horse Power (B.H.P.) were obtained:
B.H.P
10
21
30
42
470
53
W\b
107
216
285
391
440
490
Represent the above graphically and estimate B.H.P. when
JP«41b.
23. Toronto required $30,080,000 during 1920, to meet
civic expenses. This was obtained as follows:
General taxes $13,074,312
School taxes 6,396,788
Water rates 2,840,066
Surplus from 1919 2,415,345
Hydro 606,069
Local improvements 1,605,675
Street railway 1,098,651
Abattoir
Rentals
Licenses
City car lines . .
C.N.E
Fines
Other revenues.
130,000
186,600
113,000
445,000
100,000
150,000
917,120
Employ the circular pictograph to represent the above.
24. The following are the results obtained with a screwjack:
Load in lb
5
10
15
20*
28
30
35
40
45
Effort in lb
.172
.282
.359
.409
.578
.688
.797
.960
1.000
1.100
Friction in lb. .
19.86
27.48
46.77
62.04
75.50
83.13
Efficiency %.. .
15.4
29.9
32.6
34.6
35.1
On the same sheet of paper draw graphs of the relation
between load and effort, load and friction, load and efficiency.
Estimate the missing quantities from your graph.
GRAPHS
169
25. The results shown in the following table were obtained
experimentally from a lifting machine. Plot the two curves
connecting P and W and F and W:
Load (W) lb
5
10
15
20
25
30
35
40
Effort (P) lb
.094
.45
.81
1.17
1.53
1.88
2.61
2.97
Friction (F) lb
2.34
6.32
10.31
14.29
18.28
22.26
34.21
Estimate the missing quantities from your graph.
26. The tax rate in Toronto in 1919 was 28£ mills, divided
as follows:
General City purposes 10.89 mills
Schools 7.90 mills
Public Library 0.25 mills
Administration of Justice 2.27 mills
Street Maintenance 3 . 93 mills
War Expenditure 3 . 26 mills
Employ the circular pictograph to represent the above.
27. The increase in wages of the employees of a railway
company from 1913 to 1916, based on $1 a day, is given as
follows :
Trackmen from $1 . 15 to
Station Agents from 1.75 to
Office Clerks from 2. 10 to
Trainmen from 1 . 80 to
Machinists from 2 . 25 to
Conductors from 3. 15 to
Enginemen from 3 . 60 to
Represent these increases graphically.
28. The following table gives the edible portions of various
kinds of fish and the price per pound:
1.30
2.25
50
80
20
25
4.75
Kind
Halibut
Haddock
Whitefish
Bass
Herring
Perch
Pike
Canned
Salmon
Edible portion
in %
72
49
56
45
57
37
42
86
Price per lb....
24 .
18
20
22
16
12
18
32
Express graphically the edible portions of these various kinds
of fish that can be bought for $1.
170
MATHEMATICS FOR TECHNICAL SCHOOLS
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CHAPTER XIV.
MATHEMATICS OF THE MACHINE SHOP.
115. Machinist's Scale. A machinist's scale is made of steel
and is usually either 6" or 12" in length. There are markings
on the four edges, 8ths, 16ths, 32nds, and 64ths. In measuring
machinists prefer to " split ". a 32nd, instead of attempting
to read to 64ths.
116. Try Square. The try
square is used for testing if
surfaces are at right angles
to one another. The dia
gram illustrates its use for
testing a piece of work.
117. Calipers. Frequently
it is not possible to obtain
an accurate measurement
with the scale, for example
the outside diameter of a fig. 64
cylinder, or the inside diameter of a pipe. For such purposes
calipers are used. These are of three types — outside calipers,
inside calipers, and hermaphrodite calipers.
1
BLADE
o o
o
o o
1
<0
172
MATHEMATICS FOR TECHNICAL SCHOOLS
Outside calipers are used for taking outside dimensions as
the diameter of a cylindrical piece of work, inside calipers for
measurements such as the bore of a pipe, and hermaphrodite
calipers for finding the centre of a piece of work and for
scribing.
The calipers must be finely adjusted so that they will just
touch the sides of the work as they pass over it. Care must
also be taken to keep them at right angles to the work.
Fig. 66
In laying the calipers on the scale to find the length, place
one leg at the end of the scale and read the mark on the
scale where the other leg touches (see Figure 66).
118. Centring. Work is frequently held in a lathe between
two points called centres. In order to accommodate these,
small holes must be drilled in the ends of the work. These
holes are countersunk to the same angle as the centres,
usually 60°.
MATHEMATICS OF THE MACHINE SHOP
173
(1) Centring by hermaphrodite calipers. The calipers are
set so that the pointed leg reaches
approximately the centre of the
work. The calipers are then placed
at A, B, C and D and arcs are
described as shown. The centre of
the work will be the centre of the
figure thus obtained.
Fig. (i7
(2) Centring by the centre square.
The centre square consists of a
head and blade. The head is so
adjusted that the edge of the blade
comes across the diameter of a piece
of round stock placed in the head as
shown.
A line is drawn along the blade on
the work. The work is then turned
to some other position and another
diameter is drawn. Where these
diameters cross will be the centre.
119. Vernier. With the scale,
Figure 69, we could measure to a cer
tain degree of accuracy. If the length Fia
came between 7 and 8, we could estimate the amount, say
174
MATHEMATICS FOR TECHNICAL SCHOOLS
76. For obtaining greater accuracy in this part between the
7 and 8 a device known as a vernier is used (Pierre Vernier
—1631).
Fig. 69
In Figure 70 a second scale CD, called a vernier, is placed
alongside of the scale AB and we observe that 10 divisions
on the vernier is equal to 9 divisions on the scale. Obviously
each division on CD is T V less than a division on AB. There
fore the length between the 1 mark on AB and th.e 1 mark
on CD will be T V of a division on A B. Also the length between
the 2 mark on AB and the 2 mark on CD will be ^ of a
division on AB, and so on.
In Figure 71 the reading on AB is 2 plus a decimal. To
get the decimal part we observe that division 4 of the vernier
coincides with a division on the scale. Evidently the excess
of the reading over 2 is the difference between 4 divisions
on AB and 4 vernier divisions, which as above explained is
T % of a division on AB. Therefore the reading is 24.
MATHEMATICS OF THE MACHINE SHOP
175
For more accurate readings the vernier sometimes has 25
divisions corresponding to 24 divisions on the scale.
120. Micrometer. A micrometer is an instrument for
measuring to a greater degree of accuracy than can be measured
with a scale.
Fig. 72
The above is a representation of the micrometer for
measuring in inches, the parts being indicated.
The principle of the instrument is as follows:
The screw is threaded inside of the sleeve with 40 threads
to the inch. The thimble is attached to the end of the screw
and the work to be measured is placed between the screw
and anvil. The micrometer is then closed on the work by
turning up the thimble.
Since the screw has 40 threads to the inch, one turn of the
thimble closes the opening T ^th of an inch or 025". Each
mark on the sleeve represents one complete turn of the thimble,
therefore, four turns equals 4X T V" or
i "
Figures are placed
on the sleeve at every fourth mark, representing tenths of
an inch. The thimble is divided into 25 equal divisions so
that turning the thimble one division advances the screw ^
of i =_JL_"
U1 T¥ 10 •
To read the micrometer in the above figure:
The last number exposed on the sleeve is 4, therefore we
set down T V or 4". Between the last number exposed and
176
MATHEMATICS FOR TECHNICAL SCHOOLS
the edge of the thimble two small divisions are showing,
therefore 2X T V" or 2X 025* = 05".
The thimble has evidently turned 12 spaces from the zero
mark, therefore f of ^" = T ^"= 012".
..total reading =• 4"+ 05"+ 012" =462".
If the micrometer has a vernier, of the type described, the
divisions on the thimble can be divided, making the micrometer
read to 10,000ths.
Exercises LXIX.
— &
rrrrnT=:
o I 2 b
?r~
minimi
*^*_
C « Z 3 A S
Imlmliiiliiiliiili
Fig. 73
1. Calculate the micrometer readings in the above figures.
2. If the thimble be turned backward through 6 complete
revolutions, what decimal of an inch is the micrometer
opened?
3. Through how many turns must the thimble be moved
to open the micrometer 7"?
4. The sleeve reading is 4 and the thimble reading is 18.
What is the opening of the micrometer?
5. How many turns must the micrometer be opened to
read 458"?
6. A spindle is ground to 1345". What is the setting on
the micrometer?
7. A ball measures 864". What is the setting of the
micrometer?
8. Calculate the setting of the micrometer for ^".
9. Explain how you would set a micrometer for tAt,"
over *.
10. Calculate the setting of the micrometer for f ".
MATHEMATICS OF THE MACHINE SHOP
177
121. Vernier Caliper.
bar with a sliding jaw.
The vernier caliper consists of a
The bar is divided the same as the
t y»*AAw«y
1
456769
nlni iiilinlmlmliii
nil
mi mil iimnm
O 5 fc> 15 20 2S
* A
\ 2 3f
iMlMWlMiliHli.
Fig. 74
Fig. 74a
sleeve of the micrometer, i.e., the smallest division being ^
of an inch.
On the slidingjaw is a vernier. It is divided into 25 parts,
the total length of these parts being equal to 24 divisions on
the bar. As previously described in the case of the vernier,
the distance between say the 4th mark on the vernier and
the 4th mark on the bar will be ^ of a division on the bar.
Since a division on the bar is ^ of an inch, this distance will
be T VXirff = TTnnr of an inch.
178 MATHEMATICS FOR TECHNICAL SCHOOLS
In the preceding figure our object is to obtain the bar reading
opposite the on the vernier. The last figure showing on the
bar is 3, .'. '3". From the 3 on the bar to the last division
before the on the vernier we have three small divisions.
.'. 3 X T V = 3 X 025" = 075". To get the vernier reading
we observe that the 5 line on the vernier is exactly opposite a
line on the bar, .\ nftnr" = 005"
.'. total reading = 3" + 075" + 005" = 380".
Exercises LXX.
1. What would be the correct setting for a vernier caliper
to read 1642"?
2. A reading on the vernier caliper shows 1", 3 tenths, 2
small divisions, while the 12th division on the vernier is in
line with a beam division. What is the reading?
3. How would you set a vernier caliper to read §"?
4. What fraction of an inch is represented when the bar
shows 2 tenths, 2 small divisions, and the 6th division on the
vernier is in line with a beam division?
5. How would you set a vernier caliper to read 7645"?
6. A reading on the vernier caliper shows 3", 3 tenths, 3
small divisions, while the 8th division on the vernier is in
line with a beam division. What is the reading?
122. Cutting and Surface Speed. In the running of
machinery in the shop, the workman should know the speed
at which to run the machine in order to give the best results.
Lathes, milling machines, etc., are provided with attachments
for changing the speed. This speed depends on the kind of
material in the work, whether it is a roughing or finishing cut, etc.
In the lathe the cutting speed is the rate at which the work
passes the tool and is usually reckoned in feet per min. The
same definition would apply to the cutting speed of a planer.
In the shaper, however, it is the tool that is moving and as
a consequence the cutting speed would be the rate at which the
tool passes over the work.
MATHEMATICS OF THE MACHINE SHOP 179
Cutting Speed of a Lathe. If a piece of work is being
turned in a lathe, the tool will pass over the whole circum
ference of the work in a complete revolution. If the diameter
of the work be 7", then the circumference would be ^ X7"
= 22"=2_§'. if the lathe is making 40 revolutions per min.,
then the cutting speed of the lathe = ff X 40 = *£* = 73^' per min.
From the above we have that the cutting speed of the lathe
in feet per minute = Circumference of work in ft.XR.P.M.
(R.P.M. being contraction for "revolutions per minute").
We have here three quantities involved — the cutting speed,
the circumference of the work, and the revolutions per minute.
If we know any two of these we can find the third.
Example 1:
A piece of work 5" in diameter is to be turned in a lathe.
How many revolutions per min. should the lathe make to
give a cutting speed of 35 ft. per min.?
The circumference of the work = 5 X ^ =ip."= j^l'.
The tool travels ^ ' in 1 revolution.
It would travel 35' in yyVX t 5  revs. = 26+revs.
Example 2:
The surface speed of an emerywheel is to be 200 ft. per
min. It is belted to an arbor to run 50 R.P.M. Find the
circumference of the wheel.
In 50 revolutions a point on the surface travels 200'.
In 1 revolution £££■ = 4'.
.'. circumference of wheel = 4'.
^Z) = 48".
U — 2^ Ay — AOyy .
Exercises LXXI.
1. A piece of steel f" in diameter is turned in a lathe at
100 R.P.M. What is the cutting speed?
2. A brass rod £" in diameter is revolving at the rate of
300 R.P.M.; find the cutting speed.
180 MATHEMATICS FOR TECHNICAL SCHOOLS
3. A piece of work with a diameter of 3" is being turned
at a cutting speed of 50 ft. per min. What are the R.P.M.?
4. A 30" grinding wheel is run at 50 R.P.M. What is the
surface speed?
5. At what R.P.M. should a 50" wheel be run, for a surface
speed of 300 ft. per min.?
6. What sized wheel should be ordered to go on a spindle
running 1600 R.P.M., to give a surface speed of 4000 ft.
per min.?
7. An 8" shaft is being run to give a cutting speed of 50 ft.
per min. What are the R.P.M.?
8. A castiron pulley is machined at a cutting speed of 30
ft. per min. If the R.P.M. is 10, what is the diameter of the
pulley?
9. What would be the rim speed in ft. per min. of a fly
wheel 10' in diameter, running 75 R.P.M.?
10. How many revolutions per min. will it take to turn a
piece of tool steel 2" in diameter with a cutting speed of 40
ft. per min.?
123. Cutting Feed. In turning a piece of work in the
lathe, the feed is the number of revolutions of the work to one
inch travel of the carriage.
In drilling, the feed is the number of revolutions necessary to
cause the drill to descend 1 in.
Example 1:
How many revolutions are necessary to take one cut over
a shaft 6' in length with a feed of 30?
Length of shaft = 72".
.*. number of revolutions = 72X30 = 2160.
Example 2:
How long will be necessary to take one cut over a shaft 3'
long and 3" in diameter, with a cutting speed of 30 ft. per
min. and a feed of 34?
Circumference of work = 3X^r = 3r" = ft'
Since cutting speed is 30
.'. R.P.M. = 30h ff = 30Xff
MATHEMATICS OF THE MACHINE SHOP 181
Revs, necessary to finish the work = 36X34
.'. time required = (36 X 34) h (30 X ft) =32 [min.
On account of variations in the nature of materials used,
especially of castiron, and also in the cutting capacity of
tool steels, no fixed rule can be given for cutting speeds
and feeds. Generally, for roughing — slow speed and heavy
feed; for finishing — high speed and light feed.
Exercises LXXII.
1. How many revolutions will be necessary to take a cut
over a steel rod 8' in length with a feed of 24?
2. How long will be necessary to take a cut over a shaft
22" long and 2\" in diameter with a feed of 20 and a speed
of 30 ft. per min.?
3. A piece of work 5' in length is being turned at the rate
of 60 R.P.M. If the feed be 16, what time will be necessary
to make one complete cut?
4. A castiron pulley is 18" in diameter and has a 6" face.
If the cutting speed be 40 ft. per min. and the feed 16, how
long will it take for one cut over the work?
5. A shaft 6' long and 4" in diameter is being turned at a
cutting speed of 30 ft. per min. If the feed is 20, what
fraction of the surface will be cut over in 15 min.?
6. A drill is being fed to the work at 01" per revolution.
If it makes 40 revolutions per min., in what time will it cut
through 2" of metal?
7. A drill cuts \\" into a piece of work in 15 minutes.
If it makes 36 revolutions per min., what is the feed of the
drill?
8. A drill with a feed of 100 is making 50 revolutions per
min. In what time will it cut through 2\" of metal?
9. In 10 min. one cut is taken over a shaft 3' long and 4"
in diameter. If the feed of the machine is 21, what is the
cutting speed?
10. It takes 12 min. to take one cut over a shaft 18" long
and 3" in diameter. If the cutting speed is 40 ft. per min.,
what is the feed?
182
MATHEMATICS FOR TECHNICAL SCHOOLS
124. The Trigonometrical Ratios. It is frequently necessary
to make use of trigonometrical ratios in the machine shop.
We will merely define these ratios without giving reasons for
the names assigned.
1. The Sine of an angle
2. The Cosine of an angle
3. The Tangent of an angle
4. The Cosecant of an angle =
5. The Secant of an angle =
6. The Cotangent of an angle =
_Side Opposite
Hypotenuse
Side Adjacent
Hypotenuse
_Side Opposite
Side Adjacent'
Hypotenuse
Side Opposite'
Hypotenuse
Side Adjacent'
Side Adjacent
Side Opposite'
The contractions Sin, Cos, Tan, Cosec, Sec, Cot, are used
when writing the above.
A
Fig. 75
In the above triangle:
a . ^ AC n . BC
SmB = AB CosB = AB
SecB = ^. CotB = B
Tan B =
AC
BC
Cosec B = rr,
AC
BC AC
Tables giving the values of the trigonometrical ratios of
all angles from 0° to 90° are available.
MATHEMATICS OF THE MACHINE SHOP
183
125. Taper. The taper on a piece of conical work is the
difference in diameter for one foot of the work.
r
IZ
Fig. 76
If the work in the figure be 12" in length and the diame
ters 2\" and 2", the difference of \" in the diameters is called
the amount of taper, i.e., onehalf inch per foot.
D
Fig. 77
Taper expressed as an Angle.
In the taper above the sides DA and CB when produced
meet at E. The angle AEB is known as the angle of taper.
Assume that the length H N of the piece is 12", that the
diameter AB of the small end is 6", and the angle of taper
I AEB is 10°.
In A AMD : DM = 12 tan 5°
= 1.04988"
.'.DC = 6 +2X1 04988 = 81 (Approx.)
The amount of taper is therefore 21 in. per foot.
Kinds of Tapers:
(1) Morse Taper.
Possibly the most common taper is the Morse. It is found
chiefly on lathe spindles, drill spindles, and grinder spindles.
184
MATHEMATICS FOR TECHNICAL SCHOOLS
It is approximately f " per ft., but varies somewhat according
to the following table:
Number
Inches per Foot
•625
1
•600
2
•602
3
•602
4
•623
5
•630
6
•626
Of the above numbers 1, 2 and 3 are more commonly used.
(2) Brown and Sharpe Taper (B. & S.). The Brown and
Sharpe taper is \ in. per ft. for all sizes except No. 10,
which is 516 in. per ft.
It is the taper used on milling machine arbors, the milling
machine having been developed largely by Brown and Sharpe.
(3) Jarno Taper. The Jarno taper is 6 in. per ft. for all
sizes. It is frequently used on lathe centres.
126. Methods of Cutting Tapers on the Engine Lathe.
(1) By Means of the Compound Rest.
In cutting short tapers and bevels, this compound rest
(Fig. 78) is used, the extent of the work being limited by the
length of the compound rest screw. This attachment is used in
turning headstock centres. A graduated slide divided into
degrees permits of adjustment to any required angle.
(2) By Offsetting the Tail Stock.
When the tail centre and head centre of the lathe are in
alignment, the cutting tool moves in a line parallel to a line
MATHEMATICS OF THE MACHINE SHOP
185
connecting the two centres. If a piece of work be turned
in this position, a uniform cut will be taken throughout its
Fig. 78
length. If, however, the tail centre be moved out of align
ment with the head centre, the cut will be deeper at one end
than at the other.
The following diagram will help to make this clear:
hEAD CENTKE
1_TA\L CENTRE
Fig. 79
In the above diagram the tail centre is represented as set
over an amount x. If a piece of work be turned when the
186 MATHEMATICS FOR TECHNICAL SCHOOLS
centres are related in this way, the radius of the work at the
tail centre will be less by x than the radius of the work at the
head centre. Since the taper is the difference in diameter
between the centres, it follows that the offset of the tail stock
is onehalf this difference in diameter.
Example :
A piece of work 9" long is to be turned with a taper of \"
per foot; find the amount of offset of the tail stock.
A taper of \" in 12" = a taper of ^Xh" in 9".
= a taper of f" in 9".
As the tail stock must be set over onehalf of this amount,
the required offset is tV".
In the above method it must be kept in mind that the
amount of offset of the tail stock is onehalf the difference of
the end diameters whether the taper extends the full length
of the work or not.
Example :
A steel pin 12" long is to be tapered for 8" and turned
straight for the remaining 4". The diameter at the large end
is If" and the small end is to be 1" in diameter; find the
amount of offset.
The taper in 8" = If "  l" = f "
..taper in 12"=^ off" = "
.'.amount of offset = £§".
(3) By Means of a Taper Attachment.
Many lathes are now fitted with a taper attachment. This
is attached to the back of the lathe and is connected to the
crossfeed. A movable slide can be adjusted at various angles
to the travel of the carriage, and the crossfeed screw having
been released, the crossfeed slide will move backward and
forward according to the alignment of the slide on the taper
attachment. This method should always be used if a lathe
with a taper attachment is available.
MATHEMATICS OF THE MACHINE SHOP 187
Exercises LXXIII.
1. A piece of steel 8" long has end diameters of 1" and f\
Find the amount of taper, i.e., taper per foot.
2. A piece of work 10" long has a Jarno taper and has a
diameter at the large end of If". What is the diameter at
the small end?
3. A piece of work 6" long has a No. 1 Morse taper and has
a diameter at the small end of tk". What is the diameter
at the large end?
4. A taper pin tapers \" per foot and has end diameters
of \ w and f". What is the length of the pin?
5. A piece of work with a Jarno taper has end diameters
of A" and H". What is the length of the work?
6. A piece of work with a B. & S. taper has end diameters
of 1J* and 1^" What is the length of the work?
7. A piece of work 15" long has end diameters of \\" and
2gz". What kind of taper was used in turning?
8. A piece of work 18" long has end diameters of \\" and
2". What kind of taper was used in turning?
9. A piece of work 2f" long has end diameters of 368"
and 475"; find the amount of taper.
10. A piece of work 6" long is to have end diameters of
•45" and 7625". What kind of taper would the work have
when finished?
11. A piece of work 9" long is to have end diameters of
•3125" and 6875". What is the amount of taper?
12. A piece of work 21" long has end diameters of 875"
and 1925". What kind of taper has it?
13. Explain why the offset of the tail stock is one half
the difference of the required diameters.
14. A steel pin 1\" long is to be turned with a taper of
f " per foot. What is the offset of the tail stock?
15. The diameter at the large end of a piece of work is \\"
and the diameter at the small end 1". What is the offset of
the tail stock?
16. A taper pin is 1" in diameter at the large end and ^"
at the small end. What is the offset of the tail stock?
17. A taper gauge has end diameters of 2\" and If". If
the length of the taper is 9" and the total length 12", find the
offset of the tail stock.
188
MATHEMATICS FOR TECHNICAL SCHOOLS
18. Determine the distance that the tail stock should be
set over to cut the following:
(a) A No. Morse taper on a piece of work 9" long.
(6) A No. 1 Morse taper on a piece of work 1\" long.
(c) A Jarno taper on a piece of work 10" long.
(d) A Brown & Sharpe taper on a piece of work 18" long.
19. A piece of work 18" long is to be turned straight for
12" and the remaining 6" to be tapered. The diameter at the
large end is to be 2" and at the small end 1"; find the offset
of the tail stock.
20. A tapered piece of work is 8" long, and a Jarno taper
was turned on the piece. What is the difference in end
diameters?
21. A piece of work 20" long is to be turned to a diameter
of 3" at the centre, and to be tapered from centre to each
end with a taper of \" per foot. Determine the end diameters
and the offset of the tail stock.
22. A piece of work 12" long having a diameter at the
larger end of 6", tapers to an angle of 10°. What is the
amount of taper?
23. A piece of work 18" long and a diameter at the smaller
end of 8", tapers to an angle of 8°. What is the amount of
taper?
24. What is the angle of taper in a Morse No. 0, a Morse
No. 2, a B. & S., a Jarno?
127. Threads. A thread is formed by cutting a uniform
spiral groove around a piece of work.
aaAA
¥1_L
The diameter of a screw is the distance from the point of
a thread on one side to a point on the opposite side (outside
diameter of diagram). The inside diameter is the diameter
measured at the bottom of the groove (see diagram). The
pitch of a thread on a screw is the distance from the middle
MATHEMATICS OF THE MACHINE SHOP 189
point of one thread to the middle point of the next, measured
in a line parallel to the axis. Pitch is usually stated as the
number of threads per inch. Thus if there are 10 threads
per inch, the pitch is ^.
Stated generally : Pitch = — 55 7—7 1 : — :
A o. oj threads per inch.
To estimate the number of threads per inch, place a mark
on the scale on the point of a thread and count the number
of grooves within the inch line, or count the number of threads
and subtract 1.
The lead of a screw is the distance the screw advances in
one complete turn. In a single threaded screw the pitch is
equal to the lead. Thus if the pitch is ^, the screw will move
forward jV' in one complete revolution. In a double threaded
screw the pitch is \ the lead, in a triple threaded screw •§ the
lead, and so on.
If a screw has a righthanded thread it turns in the direction
of the hands of a clock when screwed into the nut. If a left
handed thread it will turn in the opposite direction when
screwed into the nut.
Exercises LXXIV.
1. Secure a number of different kinds of screws and find
the number of threads per inch in each.
2. What is the lead of a single threaded screw if it has
(a) 6 threads per inch, (6) 12 threads per inch, (c)15 threads
per inch?
3. A single threaded screw advances 2" in 12 turns, what
is the pitch?
4. What is the pitch of a double threaded screw if it has
12 threads per inch?
5. A jackscrew has 4 threads per inch. How far does it
move in \ a revolution?
6. What is the pitch of a triple threaded screw that advances
3" in 6 revolutions?
7. What is the pitch of a double threaded screw which
advances 1" in 6 revolutions?
190 MATHEMATICS FOR TECHNICAL SCHOOLS
128. Kinds of Threads:
(1) Sharp "V" Thread.
u
1
t\ ♦
<
ll\ "
1 5
M
1 *
I
11/ ?
1/ *
V
1
Fig. 81
The sharp "V" thread is a thread having its sides at an
angle of 60° to each other and being perfectly sharp at both
top and bottom. It is difficult to get a sharp "V" thread
on account of the wear on the point of the tool in cutting.
Depth of " V " thread.
In figure above the thread has a pitch of 1", then in the
triangle ABC each side is 1" in length. The depth of the
thread will be equal to BD, the altitude of the triangle.
In the rightangled triangle BCD, DB 2 = BC 2 CD 2
.. D£ 2 = l 2 (i) 2 or Z># = V!=866".
If the pitch be only §*, then since the triangle formed would
be similar to the triangle ABC of the preceding, the depth
would be  of 866"= 433". If the pitch be ^" then for like
reason the depth would be & of 866"= 0721".
Calculations for threads are usually made on the double
depth. In a thread of 1" pitch the double depth would be
2X 866" = 1732".
Since by the above the depth is proportional to the pitch,
1*732 is used as a constant for all "V" threads.
Example :
If the pitch of a "V" thread is ^" the double depth would
be T V of 1732= 1732".
Since pitch = Number of threads per inch'
. , blg , th= 1732 _ f
U P Number of threads per inch
1 732
or, for brevity, D = — ^i where D is the double depth, an(
N the number of threads per inch.
MATHEMATICS OF THE MACHINE SHOP 191
As the root diameter of the thread is the same size as the
hole to be bored for tapping the thread, it is necessary to be
able to find this double depth in selecting the size of drill.
Example:
What sized tap drill must be used for \" screw, sharp "V"
thread, having 12 threads per inch?
Root Diameter = Outside Diameter — Double Depth
1732"
~' 5 " 12
= .5"_. 1443"= 3557".
From the table of decimal equivalents ff is the next above,
therefore the correct size.
Exercises LXXV.
1. By means of the method used in the preceding find the
double depth of sharp "V" threads having 8, 12, 14, 18, 20
threads per inch. Check by formula.
2. If the double depth of a sharp "V" thread is 1443",
find the number of threads per inch.
3. If the double depth of a sharp "V" thread is 1732",
find the number of threads per inch.
4. What size of tap drill would be necessary for a " screw
with a sharp "V" thread having 9 threads per inch?
5. What size of tap drill would be necessary for a If" screw
with a sharp "V" thread having 7 threads per inch?
6. What size of tap drill would be necessary for a fa" screw
with a sharp "V" thread having 12 threads per inch?
7. If the single depth of a sharp "V" thread is 1733",
find the pitch.
8. The root diameter of a 3" bolt with a sharp " V" thread
is 25052". What is the number of threads per inch?
9. The root diameter of a bolt with a sharp "V" thread is
•7835". If it has 8 threads per inch, what is the outside
diameter?
10. What is the root diameter of a 2" bolt with a sharp
"V" thread having 4 threads per inch?
192 MATHEMATICS FOR TECHNICAL SCHOOLS
(2) The United States' Standard Thread (U.S. Std.).
Fig. 82
This thread is commonly used in machine work as it gets
over the difficulty of the sharp edges of the "V" thread.
This thread has the same triangular form as the sharp
"V" thread but is flattened at the point and bottom. This
flattened part is  of the pitch in width. As f of the height
is taken from the top and bottom the depth of the thread is
f the depth of the "V" thread. .. depth = f of 866"= 649",
Also double depth is f of 1732" = 1299".
As in the sharp "V" thread, the double depth of the U.S.
Std. for different pitches may be found by dividing the con
stant by the number of threads per inch.
1299
.'. Double Depth of U.S. Std. thread =
Also Root Diameter = Outside Diameter—
Number of threads per inch
1299
Nu mber of threads per in ch
To find the size of tap drill for a U.S. Std. thread we would
proceed as in the case of a sharp "V" thread.
Example :
What sized tap drill would be used for a f " screw, U.S. Std.
thread, 11 threads per inch?
Root Diameter = Outside Diameter — Double Depth
1 299"
= . 6 25" ff=' 5069".
From table of decimal equivalents ff is the next above and
consequently the correct size.
MATHEMATICS OF THE MACHINE SHOP
193
Exercises LXXVI.
1. What is the double depth of a U.S. Std. thread of  pitch?
2. What is the root diameter of a &" U.S. Std. threaded
screw, 12 threads per inch?
3. The root diameter of a f " U.S. Std. threaded screw is
•6201". What is the pitch?
4. The root diameter of a U.S. Std. threaded bolt is 3567".
If the pitch is , what is the outside diameter of the screw?
5. If the single depth of a U.S. Std. thread is 0491", find
the pitch?
6. If the double depth of a U.S. Std. thread is 3248",
what is the number of threads per inch?
7. The single depth of a U.S. Std. thread is 1998", what is
the number of threads per inch?
8. What sized tap drill would be used for a If" screw, U.S.
Std. thread, 7 threads to the inch?
9. What sized tap drill would be used for a If" screw,
Std. thread, 5 threads to the inch?
10. What sized tap drill would be used for a 1" screw,
Std. thread, 8 threads to the inch?
(3) Square Thread.
U.S.
U.S.
Fig. 83
The square thread is used in screws which are subjected to
heavy loads, the jackscrew being an example.
In this thread the sides are parallel, the thickness of the
tooth, the depth, and the width of the groove being all theo
retically equal. In practice, however, the width of the groove
is made slightly larger than the thickness of the thread to allow
for clearance.
The pitch — or the distance from the middle point of one
tooth to the middle point of the next — is in the square thread
equivalent to one tooth and one space.
194
MATHEMATICS FOR TECHNICAL SCHOOLS
If, as in previous cases, we take a pitch of 1", then the thick
ness of the thread will be §", the depth \*\ and the width of the
groove \".
Example :
Find the root diameter of a square thread 3" in diameter,
with a pitch of \. If the pitch is , then the depth is \" .
.'. double depth = ". .\ root diameter = 2f".
We could obtain the same result by using a formula similar
to that for preceding threads.
1
Root Diameter = Outside Diameter —
.'. Root Diameter = 3 " \" = 2f*.
(4) Acme 29° thread.
No. of threads per inch
c
P' 5 ^
■•—pitch 
♦ / \
* 1 \
a 1 \
P AM N r
/? \ /
\ g
i H
§ r V
/ z
Y T
5 ' \
* '
B
Fig. 84
This thread was designed to overcome the defects in the
square thread. It is less difficult to make and does away with
the sharp corners. Its principle use is in machine tool manu
facture, where it is used for lead screws and other service
where power is transmitted.
The angle between the threads is 29°, and theoretically the
depth of the thread is one half the pitch.
If we consider the figure to the right above — pitch 1" —
we have :
In A ABC, BC= 5" Cot 14° 30'
= 5"X386671
= 193335"
.'. 2 43f = l93335"5" = l43335"
.*. AM= •71667".
MATHEMATICS OF THE MACHINE SHOP 195
In A ADM, DM = AM tan 14° 30'
= • 71667 "X 25862
= • 185345 "
.. 2 DM or DN = 37069"
.. Width of flat at top = 3707 "
Width of flat at bottom = 3707"
Width of space at top = l" 3707"= 6293".
These are constants for all pitches. In practice to give
clearance the following measurements are used (P = pitch):
Width of flat at top = 3707 P
Width of flat at bottom = 3707 P 0052"
Width of space at top = • 6293 P
Depth of thread = \ P + • 010".
Exercises LXXVII.
1. Find the outside diameter of a screw with a square thread
which has a root diameter of 1" and a pitch of ^.
2. Find the root diameter of a square thread which has an
outside diameter of 2\" and a pitch of \.
3. A square thread has an outside diameter of 4" and an
inside diameter of 3". What is the pitch?
4. Find the root diameter of a square thread which has an
outside diameter of 3" and a pitch of \.
5. By a method similar to that employed in the preceding
for a 1" pitch, find the width of flat at top, width at bottom,
width of space at top, when the Acme 29° thread has a pitch
of \. Check by means of data furnished for 1" pitch.
6. If the depth of an Acme 29° thread is 3850", what is the
number of threads per inch?
7. If the width of flat at the top of an Acme 29° thread is
• 1853", what is the pitch?
8. If the width of space at the top of an Acme 29° thread is
•1573", what is the number of threads per inch?
9. If the width of space at the bottom of an Acme 29° is
•0566", what is the number of threads per inch?
196
MATHEMATICS FOR TECHNICAL SCHOOLS
(5) Whitworth Thread.
1 \ <
1
it
11 1 Q
u
11/ 
8
Fig. 85
This is a standard thread in England and on the Continent.
The sides form an angle of 55° with one another, while the
top and bottom are rounded. The rounded part at both top
and bottom is equal to onesixth of the total depth of triangle
above, leaving twothirds for the depth of the thread.
In above figure if AD = x, BC = radius of rounded part (r),
then AB=?+r.
o
If the pitch be 1" we have:
In &ADE, AD= 5 Cot 27° 30'
= 5"X 192098
= • 96049 " = x.
x i
6 +r '
In A ABC, Cosec 27° 30'
216568 =
6 +r
216568 r = +r
6
116568 r = f =
6
96049
6
= •16008"
.'. r=1373"
depth of thread = f X • 96049" = • 64033".
The dimensions of this thread stated in terms of the pitch
(P) are as a consequence of the above:
Depth = 64033 P.
Radius of rounded part = • 1373 P.
Example:
Find the depth and radius of curvature of a Whitworth
thread having a pitch of T V
Depth = 64033 X T V = 064033".
Radius of Curvature = • 1373 X T V= ■ 01373 *.
MATHEMATICS OF THE MACHINE SHOP
197
Exercises LXXVIII.
1. Find the depth and radius of curvature of a Whitworth
thread having a pitch of yj.
2. The depth of a Whitworth thread on a f " screw is • 064033 "
"What is the pitch and the diameter at the root?
3. A 1" screw has a Whitworth thread with a pitch of • 1250".
What is the depth of the thread and the diameter at the root?
4. A 1\" screw with a Whitworth thread has a diameter at
the root of 1067". Find depth of thread, the pitch, and radius
of curvature.
5. The depth of a Whitworth thread on a 2" screw is • 1423".
Find the number of threads per inch, the diameter at the root,
and the radius of curvature.
6. A " screw has a Whitworth thread with 16 threads per
inch. Find the depth of the thread, the diameter at the root,
and the radius of curvature.
129. Thread Cutting.
Gear Trains. One of the common ways of transmitting
motion from one point to another is by means of gear trains.
Fig. 86
The simplest form of gear train, having but two gears, is
shown in Figure 86. Gears are usually known by their number
of teeth. Thus, if / has 20 teeth it would be called a 20toothed
gear. Similarly // would be called a 60toothed gear.
198
MATHEMATICS FOR TECHNICAL SCHOOLS
If two such gears are in mesh, as above, and the motion
from 7 is transmitted to 77, 7 would be known as the driver
and 77 as the driven. As each tooth in 7 pushes along a corres
ponding tooth in 77, it follows that one revolution of 7 will
cause 77 to make only onethird of a revolution. Therefore the
shaft to which 7 is keyed will make three revolutions while the
shaft to which 77 is keyed is making one revolution. This
principle is used extensively in gear trains.
Fig. 87
In Figure 87 we have three gears in the train. It may be
necessary to insert the intermediate gear 77, either, that 7
and 777 may have the same direction, or to permit of 7 driving
777 without increasing the size of the gears. The gear 77 has
no effect on the speed ratio of 7 and 777, for when 7 moves one
tooth the same amount of motion will be transmitted to 77,
which in turn will move 777 one tooth. Since each revolution of
7 will result in onefourth of a revolution of 777, therefore the
speed ratio of 7 to 777 will be 4 to 1.
This may be stated as follows:
R.P.M. of driver _ teeth on driven 80 _ 4
R. P.M. of driven teeth on driver 20 1"
Frequently it is necessary to make such a great increase or
decrease in speed, that to accomplish it with a simple train of
MATHEMATICS OF THE MACHINE SHOP
199
gears, would necessitate too great a difference in diameters.
For this purpose a compound gear train is used.
Fig. 88
Figure 88 represents a common form of a compound gear
train, I drives II and causes a reduction of speed, /// is keyed
to the same shaft as //and therefore travels at the same speed,
/// meshes with IV, and on account of their relative number
of teeth, a further reduction of speed is effected.
If we wish to find the speed ratio of / and IV we might
proceed as follows:
Since / has 20 teeth and // 40 teeth, the speed of / is twice
that of //. Since /// has 20 teeth and / V 80 teeth, the speed
of ///, that is of //, is four times that of IV. Combining
these statements we have that the speed of / is eight times that
of IV.
The above is equivalent to the following:
R.P.M. of first driver Product of No. of teeth of all the driven
R.P.M. of last driven Product of No. of teeth of all the drivers
In above figure / and /// are the drivers and // and IV
the driven.
Substituting the values from the figure in the above relation:
R.P.M. of first driver 40X80 8
R.P.M. of last driven ~ 20X20 " 1*
200
MATHEMATICS FOR TECHNICAL SCHOOLS
Cutting a Thread. If a piece of work, on which a thread is
to be cut, is placed in a lathe, it will revolve at the same rate
as the spindle. If the spindle and lead screw turn at the same
rate, then the number of threads per inch on the work will
be the same as the number of threads per inch on the lead
screw. If it is necessary that the number of threads per inch
on the work differ from the number of threads per inch on the
lead screw, then the principle of changing the speed by in
serting gears of different sizes becomes necessary.
SPINDLE. GEAR
CHANGE STUD GE AP.
INSIDE STUD GEAP.
■• MlliWiW tW till
X LEAD SCREW
^ \CHANQE LEAP 5CBEW GEAR.
Fig. 89
Figure 89 shows the relation of gears in a simple geared
lathe.
The spindle gear turns with the spindle, and drives the inside
stud gear through the idler. The change stud gear, which is
keyed to the inside stud, transfers the motion through another
idler to the lead screw.
MATHEMATICS OF THE MACHINE SHOP 201
If the spindle gear in the above has 24 teeth, the inside gear
on the stud 24 teeth, the change stud gear 40 teeth, and the
lead screw 80 teeth, then: — Speed of spindle ff of speed of
stud. Speed of stud f§ of speed of lead screw. .*. Speed of
spindle = ftXf§ = f of speed of lead screw.
The same result may be obtained by substituting in the
formula, giving:
Revolutions of spindle _ Product of No. of teeth in driven
Revolutions of lead screw Product of No. of teeth in drivers
= 24X80 2
" 24X40" 1'
In this case if the lead screw has 6 threads per inch, then the
work would have 12 threads per inch.
Knowing the Lead of the Lathe we can find an arrangement
of gears which will give the desired number of threads per
inch on the work.
Example:
If the lead of the lathe is 8, find the necessary gears on stud
and lead screw to cut a thread of T V pitch.
In this case the lead screw will advance " in one
revolution and we want the work to advance ^ in one
revolution.
This ratio of 8 to 10 would be obtained if we placed an
8toothed gear on the stud and a 10toothed gear on the lead
screw. These gears are, however, not obtainable, but the same
ratio may be maintained if we place a 48toothed gear on the
stud and a 60toothed gear on the lead screw.
Gears furnished with a Lathe. Gears for a lathe usually
vary in size by adding the same number of teeth each time
to the gear just below. The two common sets are those
obtained by adding 4 to the one below, giving 24, 28, 32 ... . 120,
and those obtained by adding 7, giving 21, 28, 35.... 105.
This is called gear progression.
202 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises LXXIX.
1. A lead screw has 6 threads per inch. What gears must
be placed on stud and lead screw to cut 16 threads per inch?
2. Determine the change gears for cutting a ^ pitch thread
when the lead screw has a § pitch.
3. A lathe with a lead screw of £ pitch has a 24toothed gear
on the stud and a 60toothed gear on the lead screw. How
many threads will be cut on a screw when the carriage has
advanced 3 inches?
4. How many threads per inch will be cut by a lathe when
the lead screw has a 64toothed gear and the stud a 24toothed
gear, the lead screw having a £ pitch?
5. We wish to cut 24 threads per inch on a lathe with a
lead of 8 and a gear progression of 4. What gears would
be used?
6. The lead screw is  pitch, the screw to be cut j? pitch.
If there is a 24toothed gear on the stud, what gear must be
placed on the lead screw?
7. A lathe having a 72toothed gear on the lead screw and a
24toothed gear on the stud cuts 18 threads per inch. What is
the pitch of the lead screw?
8. The lead screw has a 96toothed gear and a £ pitch. What
gear on the stud will cut 24 threads per inch?
9. What gear must be used on the lead screw in order to
cut 12 threads per inch, when the lead screw has 6 threads per
inch and a 36toothed gear is used on the stud?
10. We wish to cut 14 threads per inch on a lathe with
a lead of 6 and a gear progression of 7. What gears would
be used?
Compound Gearing in the Lathe. Owing to the limited
number of gears and also to give a wider range of speeds to
those available, the principle of compound gears is used on the
lathe.
Figure 90 represents the arrangement of gears on a lathe
when compounding is necessary.
MATHEMATICS OF THE MACHINE SHOP
203
The only difference between this arrangement and that
described in the simple geared lathe is that instead of the idler
which meshes with the lead screw gear, there are two gears
keyed to the same shaft. The inside one has usually twice as
many teeth as the outside, and as a consequence causes an
SPINDLE GEAE.
1=^2
CHANGE STUPGEAg ,
OUTSIDE COM
POUNDtNQ GEAB
INSIDE STUD GEAe
■E
wii i iiwtw mwtffl* 
\ LEAD SCREW
jj frChANGE LEAP 5CREW GEAg.
Fig. 90
additional reduction of speed in the ratio of 2 to 1. By the use
of this compound the same gears which are used to cut 9,
10, 12, etc., threads on the simple geared lathe will cut 18, 20,
24, etc., threads.
Example 1:
We wish to cut 24 threads per inch on a lathe with a lead of 6.
We may use the same gears as in the example under the
simple geared lathe, i.e., a 24 on both spindle and inside stud,
a 40 on the change stud, and an 80 on the lead screw gear. If
now we place a compound between the change stud and lead
204 MATHEMATICS FOR TECHNICAL SCHOOLS
screw gear, consisting of a 72 and a 36 (see diagram), we would
have the following speed ratio:
Revolutions of spindle _ Product of No. of teeth in driven
Revolutions of lead screw Product of No. of teeth in drivers *
24X72X80 4
= 24X40X36 ~ 1'
In this arrangement the spindle will make four revolutions
when the lead screw is making one, therefore 24 revolutions
when the lead screw is making 6 revolutions.
Example 2:
We wish to cut 3 threads per inch on a lathe with a lead of 6.
In this case it is necessary for the spindle to revolve only one
half as fast as the lead screw. For this purpose we might use
the simple gear with say an 80 on the change stud and a 40 on
the lead screw. We might also use the compound gear with
equal gears on the change stud and lead screw, say 40 and 40,
and interchange the gears on the compound, i.e., have the
change stud mesh with the small gear on the compound and
the lead screw mesh with the large gear on the compound.
Then as in preceding cases:
Revolutions of spindle _ Product of No. of teeth in driven
Revolutions of lead screw ~~ Product of No. of teeth in drivers '
_ 24X36X40 _ 1 = 3
"24X72X40 2 6 
In practice the machinist reduces the method of finding the
necessary gears when compounding to the following rule:
''Write the ratio of the speed of the driving gear to the driven
gear as a fraction, divide the numerator and denominator into
two factors and multiply each pair of factors by the same number
until gears with suitable number of teeth are found. The gears
in the numerator are the driven and those in the denominator
the driving gears."
MATHEMATICS OF THE MACHINE SHOP 205
Applying this rule to the two examples above, we have
In Example 1:
24 6X4 /6X12 X _ /4X20n
6 3X2
/ DX1A / 3:XZU \
~ V3X12/ X V2X20;'
= 72 80 = 4
36 40 1'
In Example 2:
3 3X1 /3Xl2x /1X40
6 6X
1 _ / dXIA / lAt» \
1 ~ V6X12/ X V1X40J
= 36 40 = 1
72 40 2
Reduction Gears in the Headstock. Some lathes, par
ticularly those intended for cutting fine threads, have reduction
gears in the headstock. If in this case equal gears are placed
on the change stud and lead screw, the spindle does not make
the same number of revolutions as the lead screw. The ratio
of this gearing in the headstock is usually 2 to 1, so that with
equal gears on the change stud and lead screw the spindle will
turn twice as fast as the lead screw. In such lathes this must
be taken into account in figuring the necessary gears.
Cutting of Double, Triple, etc., Threads. To cut a double
thread on a screw, say 8 per inch, we would set the lathe for
cutting half that number, in this case 4. Having cut this, turn
the work onehalf of a revolution and repeat the operation.
To cut a triple thread, set the lathe for cutting onethird the
number. Having cut this, turn the work onethird of a
revolution and repeat.
Exercises LXXX.
1. A lathe has a lead screw with a lead of 4, and has a 40
toothed gear on the stud and a 90toothed gear on the lead
screw. Using a 72 and 36 as compounding gears, "how many
threads are cut?
2. We wish to cut 11 threads per inch on a lathe w r ith a
lead of 6. If the gear progression is 4, what gears would do
the work by compounding?
206 MATHEMATICS FOR TECHNICAL SCHOOLS
3. We wish to cut If threads per inch on a lathe, the lead
screw having 6 threads per inch. If the gear progression is 4,
what gears would do the work by compounding?
4. We wish to cut 64 threads per inch on a lathe with a lead
screw having 8 threads per inch. If a 24toothed gear is used
on the stud, what gears placed on compound and lead screw
would do the work?
5. A lathe has a lead of 6. If the gear progression be 7,
calculate the change gears for cutting 14 threads per inch.
6. What gears must be used to cut 12 threads per inch on a
lathe having a lead of 6, when 36 and 72 are used as com
pounding gears?
7. A special job requires 2\ threads per inch. If the lathe
has a lead of 4, what gears would do the work?
8. In a boatlifting apparatus a 12toothed gear meshes with
a 48toothed gear. Keyed to the latter is a 12toothed gear,
which meshes in turn with another 48toothed gear on the
revolving shaft. If the revolving shaft is 3 in. in diameter,
how many turns of the handle will be necessary tofraise the
boat h\ feet?
9. A lathe has 6 threads per inch on the lead screw and a 40
and 80 on the inside and outside compound respectively.
What gears must be used on stud and lead screw to cut 3
threads per inch?
10. It is desired to cut 4 threads per inch on a piece of work
The lead screw gear has 6 threads per inch, while a 36toothed
gear is placed on the stud and a 48toothed gear on the
lead screw. What arrangement of compound gears would be
suitable?
Quick Change Gears. To avoid the difficulty of having to
calculate the necessary change gears, modern lathes are
equipped with a mechanism for this purpose.
In Figures 91 and 92 this mechanism is shown.
The device is complete in one unit, and is contained in a
box which is mounted on the front of the bed where its operating
levers are convenient to the operator. The mechanism consists
essentially of a cone of gears, an intermediate shaft, and a set
of sliding gears. The tumbler gear is permanently in mesh with
a long face pinion located inside the barrel about which the
MATHEMATICS OF THE MACHINE SHOP 207
tumbler gear pivots. This gear may be tumbled into engage
ment with any of the nine gears in the cone, thus imparting
Fig. SI
Fig. 92
nine changes of speed to the intermediate shaft which is
permanently geared to the cone. It will thus be seen that
thirtysix changes are obtained with two operating levers and
without removing any of the gears.
130. Gear Calculation. In the last section the principle of
change gears as applied to the lathe was dealt with. We will
now consider some calculations pertaining to the gear itself.
208
MATHEMATICS FOR TECHNICAL SCHOOLS
In Figure 93 some of the more important terms with respect
to a spur gear are indicated.
CIRCULAR. PITCH
TH I CKNE2>5 OF TOOTH
PITCH CIRCLE
ADDENDUM
DEDENDUM
WOKWNQ DEPTH
WHOLE DEPTH
h— ROOT DIAMETER*
' PITCH DIAMETER.—
 OUTolDB DIAMETER.
Fig. 93
The Pitch Circle is the line halfway between the top and
bottom of the teeth. When two spur gears mesh, their pitch
circles are regarded as being in contact.
The Pitch Diameter is the diameter of the pitch circle.
The Diametral Pitch is the number of teeth to every inch of
pitch diameter of the gear. If the gear has 36 teeth and is
4 in. in diameter, it is said to be a 9 pitch gear.
The Circular Pitch is the distance from the centre of one
tooth to the centre of the next, measured along the pitch circle.
The Thickness of the tooth should be slightly less than the
space between the teeth to allow for clearance, but in practice
they are calculated as being equal. As a result either the
tooth or the space is onehalf the circular pitch.
MATHEMATICS OF THE MACHINE SHOP 209
Clearance must be provided at the bottom of the space
between the teeth (see diagram). It is usually ^ of the thick
ness of the tooth measured on the pitch circle.
The Addendum is the part of the tooth projecting beyond
the pitch circle. It is reckoned as a fraction of the size of the
tooth. Thus in a 12 pitch gear the addendum would be fa*.
The Dedendum is the part of the tooth between the pitch
circle and the working depth.
The addendum plus the dedendum make the working
depth of the tooth.
The Root Diameter is the diameter measured at the bottom
of the space (see diagram).
The Outside Diameter is the diameter measured at the
outside of the gear (see diagram).
Knowing the number of teeth in a gear and the diametral
pitch, to find the size of gear blank, i.e., outside diameter.
Example :
What should be the outside diameter of a gear blank for a
gear of 98 teeth and a diametral pitch of 14?
Diameter of pitch circle = ff = 7 ".
Addendum = 3^" on one side.
= \" on both sides.
. • . Outside diameter = 7 " + \ " = 7 • 1428 ".
To Find the Depth of Cut necessary in the Preceding Example.
Total depth = Addendum f Dedendum + Clearance.
Since the clearance depends on the thickness of the tooth
it will first be necessary to determine the thickness.
Number of teeth = 98.
Since there are 14 teeth for 1" of diameter there will be 14
teeth for 31416" of circumference.
.'. Circular pitch = 3 ' ^ 16 = 2244".
14
210 MATHEMATICS FOR TECHNICAL SCHOOLS
Since the circular pitch is the distance which a space and
tooth together occupy,
.'. thickness = of 2244" = 1122 ".
Since clearance = ^ of thickness,
.'. clearance in above = 01122".
.'. Total depth = ^"+^"+ 01122"= • 15407".
Exercises LXXXI.
1. The circular pitch of a gear is 3927". What is the
diametral pitch?
2. The diametral pitch is 12. Find the circular pitch.
3. Find the thickness of tooth on a 14 pitch gear.
4. Find the total depth of tooth on a 14 pitch gear.
5. Find the thickness of tooth on a 16 pitch gear.
6. Find the total depth of tooth on a 16 pitch gear.
7. Find the outside diameter of a gear blank for a 60toothed
gear, 12 pitch.
8. Find the outside diameter of a gear blank for 20 teeth
with a circular pitch of 7854".
9. What is the number of teeth on a gear 6" outside dia
meter, 12 pitch?
10. What is the number of teeth on a gear 8" outside diameter,
6 pitch?
11. What is the pitch of a gear having 63 teeth and measuring
65" outside diameter?
12. What is the distance between the centres of a pair of
gears having 72 teeth and 54 teeth respectively, 9 pitch?
131. The Milling Machine. "Milling is the process of
removing metal with rotary cutters. It is used extensively in
machine shops today for forming parts of machinery, tools,
etc., to required dimensions and shapes. A machine designed
especially for this purpose was in existence as early as 1818,
but little progress was made in the process until after the
invention of the universal milling machine in 186162 by
Joseph R. Brown of J. 11. Brown and Sharpe."
MATHEMATICS OP THE MACHINE SHOP 211
132. Cutting Speed. In determining the cutting speed of a
lathe we multiplied the circumference of the work in feet by
the number of revolutions which the work made per minute.
In the milling machine the diameter of the milling cutter
corresponds to the diameter of the work in the lathe. The
cutting speed of the milling cutter is therefore obtained by
multiplying the circumference of the cutter in feet by the number
of revolutions which it makes per minute.
Thus if a milling cutter 6" in diameter makes 60 revolutions
per min. the cutting speed = Circumference of cutter in ft.
X Revolutions per min. = ^ X^X ^ =94+ft. per min.
133. Feed. The feed on a milling machine is usually
reckoned in inches per min. As in the case of the lathe only a
general rule can be given. "In roughing, slow speed and
heavy feed using a coarsepitch cutter. In finishing, fast
speed and light feed using a finepitch cutter." In Figure 94
following, a coarsepitch cutter and a finepitch cutter are
shown:
Fig. 94
R. H. Smith in "Advanced Machine Work" gives the fol
lowing table for speeds and feeds:
Speeds.
212 MATHEMATICS FOR TECHNICAL SCHOOLS
With Carbon Steel Cutters. Cast iron — 40 ft. per min.
Machine steel — 40 ft. per min. Annealed carbon steel — 30
ft. per min. 'Brass or composition — 80 ft. per min.
With High Speed Steel Cutters. Cast iron— 80 ft. per min.
Machine steel — 80 ft. per min. Annealed carbon steel — 60 ft.
per min. Brass or composition — 160 ft. per min.
Feeds.
Feeds for milling cutters are from 002" to 250" per cutter
revolution, and depend on diameter of cutter, kind of material,
width and depth of cut, size of work and whether light or
heavy machine is used.
In order to calculate the feed it is necessary to know the
lead of the feed screw and the number of revolutions per
minute at which it is turning. Thus, if the lead of the feed
screw is \", and it is turning at the rate of 3 revolutions per
min., then the feed = "X3 = f " per min.
Exercises LXXXII.
1. A milling cutter 4" in diameter is turning at a rate of
40 R.P.M. What is the cutting speed?
2. A milling cutter 3^" in diameter is cutting at a speed of
36f ft. per min. What is the R.P.M. ?
3. A milling cutter turning at a rate of 56 R.P.M. has a
cutting speed of 60 ft. per min. What is the diameter of the
cutter?
4. A milling cutter 6" in diameter is cutting at a speed of
66 ft. per min. What is the R.P.M.?
5. A milling cutter 2" in diameter is running at 58 R.P.M.
What is the cutting speed?
6. A milling cutter turning at a rate of 30 R.P.M. has a
cutting speed of 40 ft. per min. What is the diameter of the
cutter?
7. The feed screw in a milling machine is single threaded and
has a pitch of £. If it is turned at a rate of 6 R.P.M., what is
the feed?
8. The feed screw in a milling machine has a double thread
with a pitch of \. If it is turned at a rate of 4 R.P.M., find
the feed.
MATHEMATICS OF THE MACHINE SHOP
213
9. The feed screw on a milling machine has a lead of \" .
How many R.P.M. does it make if the feed is \\" per min.?
10. The feed screw on a milling machine has a feed of \\"
per min., and is being turned at 6 R.P.M. What is the lead
of the screw?
134. Indexing. One of the purposes of the milling machine
is to cut slots or grooves in a circular piece of work at regular
intervals. It is, therefore, necessary that it should have an
attachment for dividing the circumference of the work into
equal parts. This attachment is called the dividing head.
The process of dividing the work into equal parts is called
indexing.
The metnods of indexing may be classified as — Rapid
Indexing, Plain Indexing, Differential Indexing.
Rapid Indexing permits of only a limited number of divisions
of the circumference of the work, plain indexing extends the
number of divisions, while differential indexing permits of a
still wider range.
WEX PLATE
/^INDEX SPINDLE
Fig. 95
135. In Rapid Indexing the. index plate is fastened directly
to the nose of the spindle as shown in Figure 95. This plate
usually has 24 holes and is rotated by hand to any desired
position, being held in place by a stoppin.
214
MATHEMATICS FOR TECHNICAL SCHOOLS
Assume that in Figure 96 we have a roundheaded
bolt which is required to be milled so that the head becomes
Fig. 96
square. In this case it is evident that the work must be
turned through \ of a revolution when one side of the work has
been milled and we are ready to mill the next. We would
therefore turn the index plate \ of a revolution, i.e., 6 holes.
Using this kind of indexing we may obtain any number of divi
sions which will divide evenly into 24, as two, three, four, six, etc.
iP=H}
Fi(i. 97
136. Plain Indexing. In the figure above the index spindle
is shown with a worm and wormwheel mechanism, the worm
MATHEMATICS OF THE MACHINE SHOP
215
being attached to the crank turned when indexing. The worm
wheel is keyed to the index spindle to which the work is
attached.
The principle of Plain Indexing may be seen from the
diagram above. In the majority of index heads the worm is
single threaded and the wormwheel has 40 teeth. If, therefore,
the index crank is turned one complete revolution, the worm
will make one revolution, which moves the wormwheel one
tooth or ^ of its circumference. If, therefore, we want to turn
the wormwheel, and hence the spindle to which it is attached,
one full revolution, we must turn the index crank 40 revolutions.
If we want to turn the spindle £ of a revolution, we will turn
the index crank 8 revolutions, and so on.
If now we assume that it is required to cut seven flutes
equally spaced in a reamer, we would first insert the stoppin
Fig. 98
(see diagram) and then estimate the required number of
revolutions of the index crank. In order to index for each
flute the index crank must be turned 4j£ =5f revolutions.
216 MATHEMATICS FOR TECHNICAL SCHOOLS
To accurately fix this f of a revolution the index plate is made
with a series of holes arranged in concentric circles. A sample
plate is shown (Fig. 98).
It is required to choose some circle where the total number
of holes can be divided by 7. In the plate shown the outside
row having 49 holes will suffice.
We will, therefore, turn the index crank five complete revolu
tions, afterwards turn to the 35th hole in the outside circle of
holes and insert crankpin.
Most milling machines are furnished with three index plates,
each having six index circles. The following numbers of holes
in the index circles of the three plates are used :
15 16 17 18 19 20
21 23 27 29 31 33
37 39 41 43 47 49
Exercises LXXXIII.
1. By the rapid method show how you would index for
milling a hexagonal head on a round bolt, if the index plate has
24 holes.
2. A piece of work is to have eight sides regularly spaced.
How would you index by the rapid method, if the index plate
has 24 holes?
3. What diameter must a piece be to mill square lj" across
the flats?
4. What diameter must a piece be to mill hexagonal 1\"
across the flats?
5. Twelve flutes are to be milled in a tap. How would you
index, using plain indexing, assuming that 40 turns of the
index crank are required for one turn of the spindle?
6. It is required to cut nine regularly spaced flutes in a
reamer. How would you index, assuming a ratio of 40 to 1
between index crank and spindle?
7. Assuming a ratio of 40 to 1 between index crank and
spindle, find the number of complete turns, the proper plate,
and the number of holes for indexing 15 divisions.
8. If the ratio between index crank and spindle be 60 to 1,
what indexing would be used for 21 divisions?
MATHEMATICS OF THE MACHINE SHOP
217
9. If the ratio between index crank and spindle be 40 to 1,
what indexing would be used to cut 84 teeth in a spur gear?
10. If the ratio between index crank and spindle be 40 to 1,
what indexing would be used to cut 105 teeth in a spur gear?
11. It is required to cut 85 teeth in a spur gear. How would
you index assuming a ratio of 40 to 1 between index crank and
spindle?
A
JP/HPLE
ZL^
tt_
\i
I
Unjr &
i /
— i— Ll
a
•! — I—
t_zj
■¥
I
Fig. 99
]
137. Differential Indexing. Assume that we require to
cut 101 teeth in a spur gear. If we proceed as in plain indexing
we would conclude that the index crank must turn T 4 ^ revo
lutions for milling each slot. As this fraction will not reduce
218
MATHEMATICS FOR TECHNICAL SCHOOLS
to lower terms it would be necessary to have a plate
with a circle containing 101 holes. As such a plate is not
available for all machines a different mechanism is
necessary. The method employed in such cases is called
differential indexing.
The diagrams 99 and 99a will help to explain the
method.
Fig. 99a
The index crank and index plate are shown connected with
the worm and wormwheel as in plain indexing. On the outer
end of the spindle the gear a is fastened. To an adjustable
bracket are keyed the two gears b and c. The gear b meshes
with a and the gear c with an idler i, which in turn meshes with
the gear d. Keyed to the same shaft with d is a bevel gear e,
which meshes with another bevel gear f.
MATHEMATICS OF THE MACHINE SHOP
219
If the index plate in the accompanying figure be held
stationary then for one complete turn of the spindle it will
Fig. 100
be necessary for the crankpin to pass the point marked P
40 times. Suppose, however, that the index plate is left free
to rotate and gears in the ratio of 1 to 1 be placed on the
spindle and on the worm, then when the spindle is
making one revolution the index plate makes one revolu
tion. If the gears be so arranged that the index plate turns
in the same direction as the spindle, it would only be necessary
for the crankpin to pass P 39 times in order to turn the spindle
once. If however the index plate turns in the opposite direction
to the spindle, it is evident that the crankpin must pass P 41
times in order to turn the spindle once. This principle of
having the index plate connected with gearing so that the crank
makes other than Jfi turns for one complete turn of the spindle
is the special feature of differential indexing.
We will now return to the difficulty of cutting 101 teeth in a
spur gear. Since our object is to turn the spindle y^y of a
revolution, and 101 is not a multiple of any of the numbers on
220 MATHEMATICS FOR TECHNICAL SCHOOLS
the index plates, we will select a number on either side of 101
which is a multiple of some of the numbers. It is readily seen
that 20 is a multiple of 100 and also that ^X 100 = 40. If,
therefore, we make 100 moves of 8 holes each on the 20 hole
circle we will turn the worm 40 revolutions.
If 101 such moves be made we would have ^X101 = 40f
rev. of worm. This is f of a revolution too many, which may
be offset by moving the index plate in the opposite direction to
the spindle by suitable gears. Splitting this ratio into two
parts we have f=fX.
Since we cannot multiply these fractions by any numbers
which will give gears in stock, we write —
2_2 X 3
= (3 X 24) X (5 X 8)
= 48 24
~72 X 40*
The 48 and 24 will be placed on the drivers, i.e., a and c, the
40 and 70 on the driven, i.e., b and d. It will be necessary to
place one idler in the train, as in diagram, in order that the
index plate may turn in the opposite direction to the spindle.
Example 2:
Required to cut 83 teeth in a spur gear.
Our object here is to turn the spindle ^ of a revolution.
Since 83 is not a multiple of any of the numbers on the index
plates, we will select a number on either side of 83 which is a
multiple of some of the numbers. Thus we observe that 16
is a multiple of 80 and also that ^X 80 = 40. If, therefore, we
make 80 moves of 8 holes each on the 16 hole circle, we will
turn the worm 40 revolutions.
If 83 such moves be made we would have ^X83=4l£
rev. of the worm. As this is 1^ revolutions too many, the
index plate must move opposite to the spindle.
MATHEMATICS OF THE MACHINE SHOP 221
As gears in the ratio of 3 to 2 are in stock, it is not necessary
to split the ratio f , but write f = ff .
Here the compound would be removed and the gears on
worm and spindle connected by means of two idlers. The
48 gear would go on the driver, i.e., on a and the 32 gear on the
driven, i.e., on d.
Example 3 :
Required to cut 137 teeth in a spur gear.
By trying different combinations as in the preceding cases
we find that ^X 140 = 40.
If, therefore, we make 140 moves of 6 holes each on the 21
hole circle, we will turn the worm 40 revolutions.
If 137 such moves be made we would have £ T X 137 = 39^
rev. of worm.
This lacks ^f of a revolution, which may be offset by moving
the index plate in the same direction as the spindle by suitable
gears. As in the preceding case it is not necessary to split
the ratio but use gears in the ratio of 6 to 7, i.e., 24 and 28.
The 24 will go on the driver, i.e., on a, and the 28 on the driven
i.e., on d. One idler would be inserted to connect the worm and
spindle.
Note. — When a simple gearing is used the number of idlers
depends on whether the index plate is to turn in the same or
opposite direction to that of the spindle. If in the same
direction one idler will be inserted, if in the opposite direction
two idlers.
To obviate the necessity of working out these gears, tables
are available giving the necessary gears for all required divisions
of the work.
138. Cutting Spirals. If the gears that drive the shaft
carrying the worm gear be connected with the feed screw, then
as the table advances the spindle will rotate. This will pro
duce a spiral cut in the work, such as may be seen in a spiral
reamer or a twist drill.
222 MATHEMATICS FOR TECHNICAL SCHOOLS
The gears for this purpose are shown in the following diagram:
ZnoGEAfcOh
GEAROKbCREW
Fig. 101
The four change gears are indicated in the above figure.
The screw gear and the first stud gear are the drivers, the others
MATHEMATICS OF THE MACHINE SHOP 223
being the driven. By using different combinations of change
gears the ratio of the lengthwise motion of the table to the
rotary motion of the spindle can be varied.
139. Lead of the Machine. If the feed screw of the table
has 4 threads to the inch, and 40 revolutions of the worm are
necessary for one revolution of the spindle, then, if change
gears of equal diameters are used, the work will make one
revolution while the table advances 10 in. The lead of the
machine is therefore 10 in.
Some machines have feed screws with other than 4 threads
to the inch, but the same principle as the above will give the
lead of the machine.
140. Calculating Change Gears. In calculating gears for
screw cutting we had the following proportion:
Product of teeth in driven _ Revolutions of spindle
Product of teeth in drivers Revolutions of feed screw
In a similar way we will now have:
Product of teeth in driven _ Lead of spiral
Product of teeth in drivers Lead of feed screw
Example 1:
Find the change gears for cutting a spiral with a lead of 20",
when the lead of the machine is 10 ".
tj Lead of spiral _ 20 _ 2
' Lead of machine 10 1'
Since four gears are used we split the ratio thus:
1 1 A 1 Vl X 32> ,X Vl X 24>'~32 X
24
24
We will, therefore, place 64 and 24 on the worm and 2nd
stud respectively, and 24 and 32 on the 1st stud and feed screw
respectively.
224 MATHEMATICS FOR TECHNICAL SCHOOLS
Example 2:
Find the change gears for cutting a spiral with a lead of
8333", when the lead of the machine is 10".
Lead of spiral _8^_5
Lead of machine 10 6'
5 5 1
Splitting the ratio = =  X ^.
o £ o
_/ 5X20 \ / 1X24 \
V2X20/ X V3X247
= 100 24
~ 40 72'
We will, therefore, place 100 and 24 on the worm and 2nd
stud respectively, and 72 and 40 on the feed screw and 1st
stud respectively.
141. Position of Table in Cutting Spirals. In order that the
cutter may have clearance in cutting the groove, it is necessary
that the table of the machine should be set at an angle. This
angle depends on two things: — The lead of the spiral and the
diameter of the work to be milled. This angle may be deter
mined either graphically or by calculation.
Fig. 102
In the figure ABC is a rightangled triangle in which BC is
equal to the lead of the spiral and AC the circumference of the
work. The angle ABC will then be the required angle.
MATHEMATICS OF THE MACHINE SHOP 225
Finding the angle by calculation is however a more accurate
method, thus:
. . Circumference of work
Tangent of required angle = Lead
From trigonometrical tables this angle can readily be found.
Example :
Find the angle at which the table must be set in milling a
twist drill 1" in diameter, lead 868".
3 1416X1
If 6 be the required angle, then tan 6 = — ' — = 36193
.. = 19° 54' = 20° (approx.)
Tables are available giving the proper gears and angle of
the table for all necessary cases.
Exercises LXXXIV.
1. If the ratio between the worm and spindle of a dividing
head is 40 to 1, find the differential indexing for the following
divisions:— 83, 99, 111, 139, 159, 161, 171, 238, 269, 351.
Verify from table.
2. What is the lead of a milling machine if the feed screw
has a lead of \" and the ratio of worm to spindle is 60 to 1?
3. If the lead of a milling machine is 10", calculate the change
gears for cutting spirals with the following leads: — 9", 1067",
1200", 1667", 2200", 3056", 4000", 5500", 6482", 8".
Verify from table.
4. The following change gears were used in cutting a
spiral, on worm 72, on 1st stud 24, on 2nd stud 24, on screw
gear 48. If the lead of the machine was 10", what was the
lead of the spiral?
5. The following change gears were used in cutting a
spiral, on worm 64, on 1st stud 24, on 2nd stud 32, on screw
gear 40. If the lead of the machine was 10", what was the
lead of the spiral?
6. The following change gears were used in cutting a
spiral, on worm 86, on 1st stud 24, on 2nd stud 24, on screw
gear 40. If the lead of the machine was 10", what was the
lead of the spiral?
7. A spiral with a lead of 792" is to be cut on a gear blank
with a pitch diameter of 3"; find the angle for setting the table.
226 MATHEMATICS FOR TECHNICAL SCHOOLS
8. A spiral with a lead of 934" is to be cut on a twist drill
with a diameter of \\"; find the angle for setting the table.
9. In milling a twist drill the table is set at an angle of
15°, and the lead of the spiral is 11 724 ". Find the diameter
of the drill.
10. In milling a twist drill the table is set at an angle of
17° 30', and the diameter of the drill is If*. Find the lead
of the spiral.
Review Exercises LXXXV
1. What is the lead on a doublethreaded screw of \ pitch?
2. A screw with a triple thread has a lead of 1". What is
the pitch?
3. How many revolutions must be made with a double
threaded screw, with a pitch of j^, so that it may advance 2"?
4. A sharp "V" thread with a pitch of ^, makes 6 turns to
the inch. How is it threaded? What is the double depth
of the thread?
5. If the double depth of a sharp "V" thread is 1924",
what is the number of threads per inch?
6. A If" bolt with a sharp "V" thread has a diameter at
the root of 1 • 2784". What is the depth of the thread? What
is the pitch?
7. How long will be necessary to take a cut over a shaft
3' long and 2" in diameter with a feed of 18 and a speed of 36
ft. per min?
8. A drill cuts f " into a piece of work in 10 min. If it
makes 40 revolutions per min., what is the feed of the drill?
9. A piece of work 6" long is to have end diameters of
•7432" and 6182"; find the amount of taper and also the angle
of taper.
10. A screw with a U.S. Std. thread has 16 threads to the
inch. It has a diameter at the root of 2936". What is the
diameter of the screw?
11. A 6" screw with a Whitworth thread has a pitch of f.
What is the root diameter of the thread?
12. The width of the flat at the top of an Acme 29° thread
is 0371". What is the pitch?
13. A lathe has an 84toothed gear on the lead screw and
the pitch of the lead screw is £. What gear on the stud will
cut 18 threads per inch, simple gearing?
MATHEMATICS OF THE MACHINE SHOP 227
14. A simple geared lathe having a 96toothed gear on the
lead screw and a 32 on the change stud cuts 24 threads per
inch. What is the pitch of the lead screw?
15. The lead screw on a lathe is j pitch, the screw to be
cut ^ pitch, and the change stud gear has 24 teeth. If the lathe
be simple geared what gear must be placed on the lead screw?
16. If the lathe in the preceding question had reduction
gears in the headstock in the ratio of 1 to 2, what gear would
be necessary on the lead screw?
17. In a lathe with a lead screw of ^ pitch a 40toothed
gear is placed on the change stud and a 75 on the lead screw.
If the lathe be simple geared how many threads per inch will
be cut on the screw?
18. How many threads per inch will be cut when the lead
screw gear is 100, the change stud 60, the outside compound
24, the inside compound 48, and the pitch of the lead
screw ?
19. A lathe with a 60toothed gear on the change stud and
a 40 on the lead screw, has 6 threads per inch on the lead screw.
What compound gears are used in cutting 2 threads per inch?
20. What should be the outside diameter of a gear blank
for a gear of 24 teeth and a pitch diameter of 4?
< 21. Two gears in mesh have 66, teeth and 88 teeth respec
tively. If they are 14 pitch gears, what is the distance between
their centres?
22. A milling cutter turning at the rate of 40 R.P.M. has
a cutting speed of 40 ft. per min. What is the diameter of
the cutter?
23. What is the lead of a milling machine if the feed screw
has a lead of £ and the ratio of worm to spindle is 40 to 1 ?
24. Find the change gears for cutting a spiral with a lead
of 164", when the lead of the machine is 10".
25. A spiral with a lead of 7 5 "is to be cut on a twist drill
with a diameter of \\"\ find the angle for setting the table.
CHAPTER XV.
LOGARITHMS.
142. If we wish to multiply 100 by 1000 we may do so in
either of two ways :
(1) 100X1000=100000
(2) 100 = 10 2 and 1000 = 10 3
/. 100 X 1000 =10 2 X10 3 = 10 5 = 100000.
In the second method we observe that the product is obtained
by adding the exponents of the powers of 10 which equal 100
and 1000.
If then we had numbers expressed as powers of 10, it would
be possible to multiply them together by adding their
exponents.
Thus, if we wished to multiply 23 by 432 we might do so
by addition of the exponents of the powers of 10 which equal
23 and 432.
We are here met by two difficulties.
(1) What powers of 10 equal 23 and 432 ?
(2) What number is represented by 10 when raised to
the sum of these two powers ?
Let us consider the following set of numbers:
(1) 10, (2) 25, (3) 100, (4) 365, (5) 1000, (6) 7628, (7) 10000.
We know that in (1) 10 = 10!, and that in (3) 100 = 10 2 .
Now in (2) 25 is greater than 10, or 10 1 , and loss than 100,
or 10 2 , therefore 25 = 10 1+adec,mal 
Again in (4) 365 is greater than 100, or 10 2 , and less than
1000, or 10 3 , therefore 365 = 10 2+adeclmal 
Further in (6) 7628 is greater than 1000, or 10 3 , and less
than 10000, or 10 4 , therefore 7628 = 10 3+adeclmal 
228
LOGARITHMS 229
Tables have been worked out giving the decimal parts of
the powers of 10 in the above.
Thus, from the tables 25 = 10 1 3979i 
365 = 10 256229 
7628 = 10 3  88241 
This exponent of the power to which we must raise 10 to
give the number is called the logarithm of the number.
Thus, logarithm of 25 = 139794.
logarithm of 365 = 256229.
logarithm of 7628 = 388241.
In this system — called the Briggs' System — the base is
10, and all numbers are considered as powers of 10.
The contraction "log" is used instead of logarithm.
143. Characteristic and Mantissa.
In25=10 139794 '
the 1 in the exponent is called the Characteristic and the
•39794 the Mantissa. The Mantissa is always positive.
Characteristic written at sight.
In the above set of numbers we observe that 25 which
is greater than 10 and less than 100, has 1 for its characteristic;
that 365 which is greater than 100 and less than 1000 has 2
for its characteristic; that 7628 which is greater than 1000
and less than 10,000, has 3 for its characteristic. We, there
fore, infer that the characteristic of the logarithm of any number
greater than 1 is one less than the number of integral figures
in the number.
144. How to find the Logarithm of a number from the
Tables.
Find the log of 36.
As previously explained we at once write down the charac
teristic, 1.
230 MATHEMATICS FOR TECHNICAL SCHOOLS
To get the decimal part we go down the lefthand column
to 36, then along the horizontal row to the right, and under
the vertical column headed 0, we read 55630,
..log 36 = 155630.
Find the log of 365.
The characteristic here is 2.
To get the decimal part we go down the lefthand column
to 36 as before, then along the horizontal row to the right,
and under the vertical column headed 5, we read • 56229.
..log 365 = 256229.
Find the log of 3658.
The characteristic here is 3.
To get the decimal part we proceed as in the last case,
giving 356229. To make the adjustment for the 8, we follow
the same horizontal row out to the mean differences. In the
vertical column headed 8, we read 95. This we add to 3 • 56229,
giving the log of 3658 = 356229
95
356324
Find the log of 36587.
The characteristic here is 4.
To get the decimal part we proceed as in the last example,
giving 4 • 56324. To make the adjustment for the 7, we observe
that in the same horizontal row, under 7 in mean differences, we
have 83. Since the 7 is in the fifth place, it has only onetenth the
value that it would have in the fourth place, therefore we move
the 83 one place to the right before adding, thus 4 • 56324
83
4563323
.*. log 36587=4563323 = 456332 to 5 places.
LOGARITHMS 231
145. Position of Decimal Point. Since the division of a
number by 10 or 100 is made by moving the decimal place to
the left, the position of the decimal affects the characteristic
only.
Thus, log 3658 =356324.
log 365 8 =256324.
log 3658 =156324.
log 3658 = 056324.
146. Knowing the Logarithm of a Number to find the
Number.
Tables, called antilogarithms, have been worked out which
enable us to find a number if we know its logarithm.
If log x = 234563, find x.
Since only mantissas are recorded in the tables, the charac
teristic 2 has no bearing on what we look up, but only serves
to fix the decimal place in the result.
Proceeding with 34563 in antilogarithms, just as outlined
in finding a logarithm, we have 22131
31
15
221635
Since the characteristic of the logarithm of the required
number is 2, it must have three figures in the integral part,
.'. x = 221 635 = 221 64 to 5 figures.
Returning to the difficulty raised when we wished to multiply
23 by 432 we have:
23 = 10 136173 
432= 10 2  63548 
..23X432 = io 1  3617 ^ 2  63548 = lO 3 " 72 ^ 9936 03 = 9936 to 5
figures.
In practice the base 10 is not written down, but only the
exponents.
232 MATHEMATICS FOR TECHNICAL SCHOOLS
Example :
Find the value of 45236X31341.
log 45236 = 165514 log 3 1341 = 149554
29 • 55
57 14
1655487 1496104
Sum of logs = 1655487
1496104
3151591=315159 to 5 places.
Antilog 315159= 14158
16
30
141770
.. 45 236X31 341 = 1417 7 to 5 figures.
Exercises LXXXVI.
Employ logarithms to find the value of:
753X2008X1493.
14632X7849X1009.
936X4592X361X108.
899X613X76297X392
5037X23684X1009.
147. Logarithms Applied to Division.
We have learned by the foregoing that to multiply two
numbers together, we add their logarithms and find the anti
logarithm of the result.
Since division is the reverse of multiplication, we could
without further detail infer that, to divide one number by
another, we subtract their logarithms and find the antilogarithm
of the result.
1.
53X82.
6.
2.
1064X150.
7.
3.
48326X108.
8.
4.
38156X17928.
9.
5.
49375X473.
10.
LOGARITHMS 233
Thus, divide 365 by 73.
log of 365 = 256229
log of 73 = 186332
difference
= 69897
antilog of
•69897=
=49888
103
80
4 • 99990 .. 365 r 73 = 4 • 9999.
Example :
_ ... , .4321X14892
Find the value of U9>7X37 . 42
log 43 • 21 = 1 • 63548 log 148 • 92 = 2 • 17026
10 265
59
163558
2172969
Sum of logs of numbers in numerator = 3 • 808549 (a).
log 1497 = 2 17319 log 37 42 = 157287
206 23
217525 157310
Sum of logs of numbers in denominator = 3 74835 (6).
(a) (b) =3808549
374835
•060199= 06020 to 5 places.
Antilog 06020= 11482
11487
.*. result = 1 • 1487 to 5 figures.
234 MATHEMATICS FOR TECHNICAL SCHOOLS
An abbreviated arrangement of the work is as follows
log 43 • 21 = 1 • 63548 l og 149  7 = 2  173 19
10 206
log 14892 = 2 17026 log 37 42 = 157287
265 9 o
59 :_
3^08^9" 3 * 74835
subtract 3 • 74835
0060199
anti 006020 = 11482
5
11487
148. Logarithm of a Number Less than Unity.
We have the following:
• 01 = — = — = 10~ 2
100 10 2 1U
•001 =
= V7^=10
1000 10 3
• 0001 = io5oo = 1 l= 10  4
By our definition of logarithms we have from the above
log 1=1.
log 01 = 2.
log .001 = 3.
log 0001 = 4.
Consider the following set of numbers :
(D '*• (3) 01. (5) 001. (7) 0001.
(2) 06. . (4) 008. (6) 0007.
We know from the above that in (1) l^lO" 1 and that in
(3) • 01 = 10 2 .
LOGARITHMS 235
Now in (2) 06 is greater than 01, or 10~ 2 , and less than
•1 or 10" 1 , therefore 06 = i(r 2 + adeclmal 
Again in (4) 008 is greater than 001, or 10~ 3 , and less
than 01, or 10" 2 , therefore 008 = i() 3 + adeclmaI 
Further in (6) 0007 is greater than 0001, or 10 4 and less
than 001, or 1CT 3 , therefore 0007 = 10 4+adeclmaK
From observing the above results we infer:
(1) That the characteristic of the logarithm of a number less
than unity is negative.
(2) That the characteristic of the logarithm of a number less
than unity is one more than the number of zeros between the
decimal point and the first significant figure.
149. How to write the Logarithm of a Number less than
Unity.
Find log 067.
By the above the characteristic is — 2, and from the tables
the mantissa is 82607.
.'. log 067= 2+ 82607.
Since the mantissa is always positive we could not correctly
write this as — 282607, for that would imply that the whole
quantity 282607 is negative. To avoid this difficulty the
minus sign is placed immediately above the characteristic.
.'. log 067 = 282607, the characteristic being read " bar" 2.
Example 1: Divide 0432 by 82624.
log 0432 = 263548
log 82624 = 1917111
Diff. of logs = 2 63548
1917111
4 718369 = 4 71837 to 5 places.
Here note that we are subtracting the greater quantity from
the less, therefore in obtaining the 4 we use the law for algebraic
subtraction, i.e., change the sign of the lower line and add.
Antilog of 471837= 000522844= 00052284 approx.
236 MATHEMATICS FOR TECHNICAL SCHOOLS
Example 2:
to . rt , f 36215X0724
FlndtheValue0f 0027X936
log 36 • 215 = 1 • 55889 log • 0027 =343 136
log 0724 =285974 log 936 =297128
•41863 (a) 40264 (6)
(a) (6) = 41863
•40264
•01599
Antilog of • 01599 = 1 • 0374 1 = 1 • 0374 approx.
Exercises LXXXVII.
Employ logarithms to find the value of :
1. 43752H875. 47286X158X10" 3
2. • 0752 i 648. • 0728 X • 63 X 10 2
26 • 584 X 075 728 • 43 X 00625X19 .
" 8359 0946X10009
4. 408039^342308.
150. Logarithm of a Power.
From tables log 2 = • 30103.
.'. 2 = 10 30103 
(O) 2 = (\ n 30103 ) 2 = 1Q 60206.
In the above we observe that the log of 2 2 is twice the log
of 2, therefore to find the value of 2 2 , we would find the log
of 2, double it and find the antilog of the result.
Thus, log 2= 30103,
twice log 2= 60206.
Antilog 60206 = 399996 = 4 (nearly).
Again, log 3 =47712.
.. 3 = 10 47712 
. Q4 = /1Q.47712\4_ JQ1.90848.
LOGARITHMS 237
Here we observe that the log of 3 4 is four times the log of
3, therefore to find the value, of 3 4 , we would find log 3, take
four times it, and find the antilog of the result.
Thus, log 3 =47712,
four times log 3 = 1 • 90848.
Antilog 1 • 90848 = 80 • 9988 = 8 1 (nearly) .
Further log 9 = • 95424.
.. 9 =10 95424 
.'. 9i = (10 95424 )i = 10 47712 
Here we observe that log 9* is onehalf the log 9, therefore
to find the value of 9*, we would find log 9, take onehalf
of it, and find the antilog of the result.
Thus, log 9= 95424.
$ log 9= 47712.
Antilog 47712 = 300004 = 3 (nearly).
From these examples we infer: — To obtain any power of a
number multiply its logarithm by the exponent of the power and
find the antilog arithm of the result.
Example 1 :
Find value of (026) 3 .
Letx=(026) 3 .
Then by above log x = 3 log • 026.
= 3 (241497).
Here we have to multiply a logarithm by 3, the mantissa
being positive and the characteristic negative. We should
first multiply them separately, giving 6+124491, and after
wards combine giving 524491.
.'. log x = 5 24491.
x = • 0000175754 =  000017575 approx.
238 MATHEMATICS FOR TECHNICAL SCHOOLS
Example 2:
Find value of (026)*.
Let log s = (026)*.
then log x = \ log 026.
= \ (241497).
The same difficulty is presented here as in the preceding
example, only we have to divide by 3 instead of multiplying.
We, therefore, write $(241497) as £(3 + 141497), the object
being to make the negative part so that the 3 will divide it
evenly.
1(3 + 141497) =1471656.
.*. log x = 1471656 = 147166 to 5 places.
x= 296251= • 29625 approx.
Example 3:
Find the value of
Letx= (Fo5y
then log x = log 1 — 6 log 105.
= 06(02119).
= •12714.
Since the mantissa must always be positive we must now
change — • 12714 to a number having a positive mantissa.
Thus, 12714=1 + 1 12714.
= 1/87286.
..log* =187286.
x =746214= 74621 approx.
151. Solution of an Exponential Equation.
If 3 Z =148, find a:.
This is an exponential equation, the unknown quantity being
the exponent.
LOGARITHMS 239
x =
Here x log 3 = log 148
l og 148
log 3
^ 217026
X 47712
.'. log x = log 2 17026 log 47712.
= 336512 1678628.
 657884= 65788 to 5 places.
x = 4 • 54853 = 4 • 5485 approx.
Exercises LXXXVTII.
Find the cube root of the following numbers:
1 2727. 3. 00069. 5. 43772.
2. 08765. 4. 7248. 6. 92814.
Find the numerical value of:
7 /^i 2 !! 1 Q 41X0015)*
■ t25  34/ ' 9  ~im~'
10 (4643) 10 X(0348) 12
•0275 1* 1245X163
V T 0l83j ' 11. 15 3  2 
12 (19)*X(19)*X(19)*X264 2 y _1_
(•0418) 2 X(4365) 3 Xv/472 10 6 '
3 (3496) 14 X(1653) 34 1
' (258) 7 X(045) 65 10 12 *
14. V& + &.
1
15
(105) 8 (105)
20
16. ^Ux J
(106) 8 ~ 105) 15
Find the value of x in the following:
17. 13* = 432. 18. 6 x = 252. 19. 15* =5.
240 MATHEMATICS FOR TECHNICAL SCHOOLS
Employ the formula for the area of a triangle in terms of
its sides to find the area of the following traingles :
20. 364 yd., 213 yd., 265 yd.
21. 1648", 2339", 3118".
22. 2500 links, 3500 links, 4000 links (area in acres).
23. 276 chains, 195 chains, 143 chains (area in acres).
24. Find the length of the perpendicular drawn from A on
BC in the triangle ABC, if a = 700', 6 = 670', c = 5272'.
25. The sides of a triangle are 436", 518", and 624".
Find the side of an equilateral triangle of equal area.
26. Find the area of a circle whose radius is 7246".
27. A circle has a radius of 4346". Find the radius of
the concentric circle which divides the first circle into two
equal areas.
28. Find the diameter of a circle whose area is equal to
that of an equilateral triangle on a side of 18".
29. Find the number of gallons in a cubical cistern, each
side of which measures 186' (1 gal. =277274 cu. in.).
30. The water contained in a cubical cistern, each edge of
which measures 5', is found to lose by evaporation 03 of its
volume in a day. If the total loss be due entirely to evapora
tion, find how many gallons will be left in the cistern at the
end of 9 days, assuming it to be full at the outset.
CP
CHAPTER XVI.
MENSURATION OF SOLIDS.
152. We have already found the surfaces and volumes of
various rectangular solids. We will now proceed to deal with
some of the more specialized forms of solids.
If the block in Figure 103 has the dimensions indicated, we
can find the area of the sides, i.e., the lateral surface by finding
the area of each lateral face and adding the
results. Thus the area of the front and back
faces = 6"X18"X2 = 216 sq. in., the area of
the two side faces = 4"X18"X2 = 144 sq. in.,
giving a total lateral area of '360 sq. in.
The same result might have been obtained
by first finding the perimeter of the base and
multiplying this result by the height. Thus
perimeter of base = 6" + 6" + 4" + 4" = 20".
.'. lateral surface = 20" X 18" =360 sq. in.
Further in finding the volume of this solid
we multiplied together the three dimensions —
length, breadth and thickness. Thus volume
= 18"X6"X4" = 432 cu. in. The same result
might have been obtained by first finding the area of the base
and then multiplying this area by the height. Thus area of end
= 6"X4" = 24 sq. in. and volume = 24X18" = 432 cu. in.
153. The Prism. A prism is a solid whose sides are paral
lelograms and whose top and bottom are parallel to each other.
In Figure 104 we have represented a number of prisms
each complying with the conditions in the definition.
A prism is called triangular, rectangular, pentagonal, etc.,
according as the base is one or other of these polygons.
241
 — €>"—»■
Fig. 103
242
MATHEMATICS FOR TECHNICAL SCHOOLS
To find the lateral surface of any of the prisms below we
would proceed as in Figure 103, i.e., multiply the perimeter of
the base by the height. Thus if p be the perimeter of the base
and h the height, the area of the lateral surface of the
prism = ph.
a
Fig. Kit
To find the volume of any one of the above prisms we would
as in Figure 103 multiply the area of the base by the height. Thus
if b be the area of the base and h the height, the volume of
the prism = bh.
If we wish to find the area of the total surface of a prism,
we would add the areas of the two ends to the area of the lateral
surface.
Exercises LXXXIX.
1. Measure the various prisms in the laboratory. Make draw
ings in your laboratory book and find total area and volume.
2. The internal dimensions of a box, without a lid, are
length 8', breadth 3', depth 2'. Find the cost of lining it with
zinc at 40c. a sq. ft.
3. A rectangular tank, 13' 6" in length by 9' 9" in breadth,
is full of water. How many gallons of water must be drawn
off to lower the surface 1 "?
4. How many sq. ft. of metal are there in a rectangular
tank, open at the top, 12' in length 10' in breadth and 8' deep?
5. A prism whose base is a regular pentagon with a side
of 9" is 25^" in height. Find its total area and volume.
6. A rectangular tank is 11" long, 14 \* wide, and 10"
deep. Find the number of gallons it contains when filled
with water within an inch of the top. •
MENSURATION OF SOLIDS
243
154. The Cylinder. A cylinder is a solid whose lateral
surface is curved and whose bases are parallel to each other.
To find the lateral surface of a
cylinder. If we roll a cylinder on
a sheet of paper until it has made
one complete revolution, we observe
that the area of the paper touched
by the cylinder is a rectangle
whose length is equal to the cir
cumference of the cylinder, and
whose breadth is equal to the height
of the cylinder. From this experi
ment we infer that the lateral surface
of a cylinder = the circumference of
base multiplied by the height. Fig. 105
Therefore with the notation in the figure the area of the
lateral surface =2irrh
= irdh (d = diameter) .
To find the volume of the cylinder.
If the cylinder in Figure 106 be cut
into a number of triangular prisms
as indicated, we can find its volume
by adding together the volumes of the
prisms. Since the volume of a tri
angular prism is found by multiplying
the area of the base • by the height,
the volume of all the prisms, i.e., of
the cylinder, may be found by multi
plying the sum of the areas of the bases
by the height. Therefore the volume
of a cylinder = area of base multiplied
by the height, or V = rrr 2 h= 785±d 2 h.
Example :
A cylindrical tank open at top is 6'
high and has a diameter of 3'. Find (1) the cost of lining
Fig. 106
244
MATHEMATICS FOR TECHNICAL SCHOOLS
with galvanized iron at 20c. a sq. ft., (2) its capacity in
gallons.
Area of lateral surface =rX3X6 = 18T sq. ft.
Area of bottom
= tt() 2 sq. ft. «= 2 • 25 * sq. ft
.'. total area
= tt(18+225) sq.ft.
= 7r(2025) sq. ft.
.•. cost
= 7r(2025)X20 = $1272
Volume
= *"(f) 2 X6 cu. ft.
Capacity
= 7r(f) 2 x6X6232gal.
= 26431 gal
Exercises XC.
1. Measure the various cylindrical models in the labora
tory. Make drawings in your laboratory book and obtain
the lateral area and volume in each case. In the case of the
iron and steel models find their weights from knowing their
volumes. Check by weighing.
2. Fill in the omitted entries in the following cylinders:
No.
Diameter
Height
Circ. at Base
Area of Base
Lateral Area
Volume
1
5"
H"
2
8"
7"
3
1"
28 sq. ft.
4
3'
154 sq.ft.
5
V
616cu.ft.
6
8'
44'
3. Find the weight of a steel shaft 2" in diameter and 12'
long.
4. A tank car is 33£' long and 8£' in diameter. How many
gallons of oil will it contain?
5. Find the cost of painting the inside of an open cylindrical
tank 10' in diameter and 15' high at 20c. a sq. yd.
MENSURATION OF SOLIDS
245
6. A cylindrical vessel partly filled with water is 8" in
diameter. A steel crane hook is immersed in the vessel and
the surface of the water is raised 2". Find the weight of the
crane hook.
Fig. 107
155. The Hollow Cylinder. The total surface of the hollow
cylinder in Figure 107 would consist of the outside lateral
surface, the inside lateral surface, and the two rims.
The outside lateral surface = ttX8X 18 = 144 * sq. in.
The inside lateral surface = tt X 6 X 18 = 108tt sq. in.
The area of the rims = ttX7X1X2 = 14tt sq. in.
The total lateral surface = 266 ■* sq. in. =83568 sq. in.
The volume of the hollow cylinder would he the area of the
base multiplied by the height.
Area of the base, i.e., the area of the ring in Figure 107
= tX7X1 = 7tt sq. in.
.'. the volume = 7ttX 18 = 395 84 cu. in.
Exercises XCI.
1. Measure the hollow cylindrical models in the laboratory.
Make drawings in your laboratory book and calculate the total
surfaces and volumes.
2. Find the whole surface of a hollow cylindrical pipe,
open at the ends, if the length is 8", the external diameter
10" and the thickness 2".
246
MATHEMATICS FOR TECHNICAL SCHOOLS
3. An iron roller is in the shape of a hollow cylinder whose
length is 4', external diameter 2' 8" and thickness \" . Find
its weight if a cu. ft. .of iron weighs 486 lb.
4. A portion of a cylindrical steel shaft casing is \2\' in
length, \\" thick, and its external diameter is 14". Find its
weight.
5. Find the weight of a lead pipe 8' long, external diameter
8", internal diameter 7", assuming that the weight of the two
flanges is equivalent to one foot length of pipe.
6. Find the weight of a hexagonal castiron nut 1" to the
side, \" thick, inside diameter f ".
156. The Right Cone. A cone is a solid whose base is a
circle and whose sides taper uniformly to a point directly
over the base.
Fig. 108
Fig. 109
Lateral Surface of a Cone. If a piece of paper be wrapped,
without crumpling or tearing, around the lateral surface of a
cone (Figure 108) and cut along the edge of the base and the
line AB, and then folded out, the paper will be a sector of a
circle (Figure 109).
MENSURATION OF SOLIDS
247
The radius AB of this sector is equal to the slant height of
the cone and the length of the arc BD is equal to the
circumference of the base of the cone. The area of a sector
of a circle has previously been found to be equal to fare
X radius. Therefore the lateral surface of a cone = \cir
cumference of the base multiplied by the slant height, or with
the notation of the figure, the lateral surface of a cone
= 2irr X hs = *d X \».
Perform the experiment suggested above. Make drawings
and write conclusions in your laboratory book.
Fig. 110
The Volume of a Cone. Take two vessels, one conical and
the other cylindrical, having the same height and radius of
end. (Figure 110).
If we fill the conical vessel with water and empty it into the
cylindrical one, we find that it will take three fillings of the
conical vessel to fill the cylindrical one.
We, therefore, infer that when the vessels are related as
in the above illustration, the volume of the cone is onethird
that of the cylinder.
248 MATHEMATICS FOR TECHNICAL SCHOOLS
But the volume of the cylinder = area of base multiplied
by the height. Therefore the volume of a cone = area of
base X3 perpendicular height, or V = irr 2 X\h = \ir r 2 h.
Note. — Height means perpendicular height, unless otherwise
stated, but in the formula for the volume of a cone we should
state "perpendicular" height to distinguish from "slant"
height in the formula for the lateral surface.
Perform the experiment suggested above. Make drawings
and write conclusions in your laboratory book.
Example :
A conical tent has a diameter at the base of 14' and a height
of 7'.
Find (1) the number of sq. yd. of canvas in the tent.
(2) the number of cu. ft. of air space.
Slant height of cone = V7 2 +7 2 = 989'
Number of sq. yd. =7rX14X 9 ^p X£
= 2417
Airspace = 7rX7'X7Xf = 359 19 cu. ft.
Exercises XCII.
1. Measure the various conical models in the laboratory.
Make drawings in your laboratory book and calculate lateral
surfaces and volumes. Find the weights of iron models from
knowing their volumes. Check by weighing.
2. A piece of paper in the form of a circular sector, of
which the radius is 8" and the length of the arc 12", is formed
into a conical cap. Find the area of the conical surface and
the base of the cone.
3. Find the weight of a castiron cone, diameter of base
7" and height 15".
4. Find the weight of petroleum in a conical vessel, diameter
of the base 14", height 10", specific gravity of petroleum 87.
5. The interior of a building is in the form of a cylinder of
20' radius and 15' in height. A cone surmounts it, radius of
base 20' and height 8'. Find (a) the cost of painting the
interior at 20c. a sq. yd., making no allowance for openings,
(6) cubic feet of air space in the building.
MENSURATION OF SOLIDS
249
6. How many yards of canvas 27" wide will be required to
make a conical tent 7 yd. in diameter and 10' high?
157. The Pyramid. A pyramid is a solid whose sides are
triangles and whose base is any figure bounded by straight
lines.
In Figure 111 we have the simplest
type of a right pyramid, the base
being a square.
Lateral Surface of a Pyramid. In
Figure 111 the lateral surface consists
of four equal isosceles triangles.
Area of ACD = CDXhAE.
.*. area of four faces = 4 times
CDXhAE.
But 4 times CD = perimeter of base, and AE = slant height
of pyramid.
.'. lateral surface of pyramid = perimeter of base X  slant height.
= \ V s (V = perimeter, 5 = slant ht.).
It may readily be shown that this formula holds where the
base is any regular polygon.
Fig. 112
Volume of a Pyramid. Take two vessels one a square
pyramid and the other a rectangular prism of the same height
250 MATHEMATICS FOR TECHNICAL SCHOOLS
and area of end as in Figure 112. If we fill the pyramidal
vessel with sand and empty it into the prism, we find that
it takes three fillings of the pyramid to fill the prism. We
therefore infer that, when the vessels are related as above,
the volume of the pyramid is onethird that of the prism.
The volume of prism = Area of base multiplied by the
height.
.*. volume of pyramid = area of baseX^perp.height.
or V = \Ah (A = area of base, h = height).
Example:
A granite pyramid 12' high stands on a square base 10'
to the side. Find (1) cost of polishing the lateral surface at
10c. a sq. ft. (2) weight, if 1 cu. ft. weighs 165 lb.
Slant height = Vl2 2 +5 2 = 13'
Lateral surface =4X10X ^ sq. ft.
Cost of polishing = 4 X10X 1 / X ^ =$26.00
Volume = 1X10X10X12
= 400 cu. ft.
Weight =165X400
= 66,000 lb.
Exercises XCIII.
1. Measure the various pyramidal models in the laboratory.
Make drawings and calculate lateral surfaces and volumes.
2. What is the weight of a castiron pyramid with a square
base 6" to a side and a height of 10"?
3. Find the total surface of a hexagonal pyramid with a
base 3" to the side and a slant height of 12". Find its weight
if made of castiron.
4. Find the number of cu. ft. of air space in a hexagonal
room, each side of which is 12', and its height 18', which is
furnished above with a pyramidal roof 9' high. Find also
the cost of painting the interior at 25c. a sq. yd., making no
allowance for openings.
5. A pyramid has a square base each side of which is 248",
and the pyramid has equilateral triangles for sides. Find
its volume.
MENSURATION OF SOLIDS
251
Frustum of Cone or Pyramid. A frustum of a cone or
pyramid is the part contained between the base and a plane
drawn parallel to it.
Lateral Surface of Frustum
of Cone. aABb in Figure 113
may be considered as a tra
pezium, ab and AB being the
parallel sides and either aA
or bB representing the per
pendicular distance between
the parallel sides.
If we consider this figure
as being bent around until
a coincides with b and A
with B, it would take the
form of a frustum of a cone,
the parallel sides of the trapezium becoming the circumferences
of the ends and the perpendicular distance between the
parallels becoming the slant height of the frustum.
Fig. 113
Fig. 1U
Since area of trapezium = Sum of parallel sides multiplied
by  perp. distance between them.
252
MATHEMATICS FOR TECHNICAL SCHOOLS
.'. lateral surface of frustum of cone = sum of circumferences
of ends X§ slant ht.
or, Lateral surface = \ (C+c)S.
= tt (R+r)S.
Lateral Surface of Frustum of Pyramid. If we consider the
frustum of a pyramid in
Figure 115, we observe that
its lateral surface is made
up of four equal trapeziums.
Area of the face cCDd =
{cd+CD)\qQ.
.'. area of the four faces =
±(cd + CD)\qQ.
But 4 (cd+CZ))= Sum of
perimeters of ends and qQ = slant height of frustum.
.*. lateral surfaces of frustum
of / pyramid=sum of perimeters
of ends X \ slant ht. = \
(Pi + Pa) S (P x and P 3 being
perimeters and S slant height).
Volume of Frustum of Cone
or Pyramid. In Figure 116
from similar triangles Ocb and
x _ r
x~+h~R'
.'. x\h =
Fig. 115
OCB we have
hr
 • x= »
K— r
Volume of whole cone
= l T /?2 IlR
1 ' Rr
Volume of small cone
, , hr
volume of frustum = l^h
Fig. 116
R 3 r
Rr
i7rh{R 2 + Rr+r 2 }.
MENSURATION OF SOLIDS
253
If A represents area of large end and a area of small end,
then A — R 2 and a = Trr 2 .
.". volume of frustum ^AA ha+s/Aa}
o
Work through a similar proof to show that the volume of a
frustum of a pyramid is the same as the above.
Example:
A vessel in the form of a
frustum of a cone has the
following dimensions: Depth
16", diameter of large end
12", diameter of small end
8". Find (a) its lateral
surface (b) its capacity in
gallons.
In the rt.angled triangle
ABC, ^C=Vl6 2 +2 2
= 1612".
Lateral surface
= (7rl2+7r8)
1612
= 20^x806 = 50643 sq. in.
Volume
= JUL {7r6 2 + ir4 2 +\A6 2 ir4 2 }
= V L {^6 2 +7r4 2 +7r6X4}
= ^7r{6 2 +4 2 +24}
Fig. 117
=  1 / 7T 76 cu. in
167
76
Capacity in gallons = ^ X 2?
459.
Exercises XCIV.
1. Measure the frustum models in the laboratory. Make
drawings in your laboratory book and calculate lateral sur
faces and volumes.
In the case of the iron and steel models find weights from
knowing their volumes. Check by weighing.
254
MATHEMATICS FOR TECHNICAL SCHOOLS
2. Find the lateral surface of the frustum of a pyramid,
perpendicular height 6", and a square base, side 6", the side
of the upper square being 1".
3. A tapered piece of castiron 2' long is 8" in diameter at
one end and 12" in diameter at the other; find its weight.
4. A piece of steel 16" long is 4" in diameter at the large
end. The taper is a Brown and Sharpe — \" to 1'; find its
weight.
5. Find the volume of a steel pin 8" long, diameter of small
end 2", the taper being a No. Morse — f " to 1'.
6. Two buckets, one cylindrical of 7" diameter, the other
a frustum of a cone with the diameters of its ends 6" and 8"
are of the same depth, 9". Find the difference in their volume.
158. The Sphere. A sphere is the geometrical name for a
round or ballshaped solid.
Fig. 119
Area of Surface of Sphere. It has been found by measure
ment that the surface of a sphere is equal to the lateral
surface of a cylinder of the same diameter and height, as
illustrated in Figure 119.
The circumference of the cylinder is 2nr and its height 2r,
hence area of surface of sphere = 27rrX2r = 4tjt 2 .
MENSURATION OF SOLIDS
255
An approximate idea of this relation may be obtained by
the following experiment :
Insert a nail in the centre of the curved surface of a hemi
sphere. Fasten the end of a cord to this nail and wrap it with
the object of completely covering the curved surface.
Next insert a nail in the centre of the flat face and wrap the
cord with the object of completely covering the flat surface.
It may then be observed that the length of cord required to
cover the flat surface is only half that required to cover the
curved surface.
But area of flat surface =7rr 2 .
.*. area of curved surface =27rr 2 .
.\ total surface of sphere = 47rr 2 .
Since 4tit 2 = 7r(2r) 2 .\ area of surface of sphere in terms of
diameter = 7rD 2 .
Volume of Sphere. The sphere may be considered as made
up of a number of pyramids, as in
dicated in Figure 120, whose bases
together form the surface of the sphere
and whose apexes meet at the centre.
Since the volume of a pyramid equals
area of base multiplied by ^ perp. ht.
( = radius of sphere), .\ volume of sphere
= Surface X i radius.
= 4;rr 2 X r.
Fig. 120
*
Volume in terms of diameter = 5236Z) 3 .
Example :
Find the surface and weight of a castiron ball, radius 5".
Surface = 4tt5 2 = 314 16 sq. in.
Volume =7r5 3 = f Ti125 cu. in.
Weight =7rl25X26 = 13613 1b.
256 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises XCV.
1. Measure the spherical models in the laboratory. Cal
culate areas and volumes.
2. Secure cylinder and sphere related as in Figure 119.
After placing sphere in cylinder, fill the remaining space with
sand. Remove sphere and replace the sand. By estimating
the part of the cylinder now occupied by the sand derive the
formula for the volume of the sphere.
3. Find the number of yards of material, 27" wide, necessary
to make a spherical balloon 12' in diameter.
4. Find the weight of a ball composed of a castiron sphere
4" in diameter, covered with a layer of lead 1" thick.
5. Find the weight of a hollow castiron sphere, internal
diameter 2\", thickness \" .
6. How many ounces of nickel would be used in plating
a ball 3" in diameter, to a depth of ^"? (1 cu. in. nickel
weighs 514 oz.).
Segment of a Sphere. A segment of a sphere is the part
cut off from a sphere by a plane.
Lateral Surface of a Segment. If we roll a sphere on a sheet of
paper, and keep in mind that the area
of the surface is equal to that of a
cylinder with radius of base equal to the
radius of the sphere and height equal to
the diameter of the sphere, we could infer
fig. 121 that the surface traced out by any
segment is equal in area to a rectangle
having the circumference of the sphere for length and the
height of the segment for width.
.'. lateral surface of segment
= 2ttRx1i.
= 2ttM.
The volume of the segment in Figure 121 (less than a hemi
vJir^ ttJi^
sphere) is given by the formula : V = —= — H jr~ •
Z o
Zone of Sphere. A zone of a sphere is the part cut off from
the sphere between two parallel planes.
MENSURATION OF SOLIDS
257
Lateral Surface of Zone
the area traced out when
paper would be equal in area to a
rectangle having the circumference of the
sphere for length, and the thickness of
the zone for breadth.
.". lateral surface of zone = 2irRh.
The volume of the zone in Figure 122
is given by the formula:
As in the segment of a sphere
rolled on the
S03JZ^
1*
f T T \
1=
z=zfi53T—
Fig. 122
Sector of Sphere.
Fig. 123
A sector of a sphere consists of a segment
and a cone whose bases are coincident,
the apex of the cone being at the
centre of the sphere.
The surface of the sector would be
equal to the surface of the segment
plus the surface of the cone.
The volume of the sector
is given
by the formula :
V = l{r>(h+2R) +/* 3 }
159. Bead. Volume
sphere is pierced by a
remaining
cylindrical
when
solid.
Volume of bead as
bead).
shown =  n (h = ht. of
6
Fig. 124
Exercises XCVI.
1. The silk covering of an umbrella forms a portion of a
sphere of 3^' in diameter, the area of the silk being 14 1 sq.
ft. Find the area sheltered from vertical rain when the handle
is held upright.
2. A sphere of diameter 24' is placed so that its centre is
37' distant from the observer's eye. Find the area of that
part of the sphere's surface that is visible to the observer.
258
MATHEMATICS FOR TECHNICAL SCHOOLS
3. A cylindrical tank is 8' long and 2\' in diameter. The
ends are spherical segments whose centre of curvature projects
6" beyond the base of the segment. Find the total surface
and volume of the tank.
4. If the diameters of two circles of a spherical zone are
12 " and 4", and the thickness of the zone 6", find its total
surface and volume.
5. In the sector of a sphere of radius 10", the height of the
segment is 4"; find the volume of the sector.
160. Solid Ring.
Ftg. 125
Examples of solid rings are found in
anchor rings, curtain rings, etc. It will
be observed that any crosssection of
such a ring will be a circle, so it may be
considered as a cylinder bent around in a
circular arc until the ends meet. The
mean length of the cylinder will be 2irR.
.'. with notation of figure:
Surface = 2 rrr x 2>rR = 4:7r 2 rR.
Volume = Trr 2 X 2*R = 2* 2 r 2 R.
161. Wedge. A wedge, as shown, is a solid contained by
five plane faces; the base is a rectangle, the two ends are
triangles, and the two remaining
faces are trapeziums having a
common side, called the edge,
which is parallel to the base.
The surface of the wedge is
found by calculating separately
the area of each of the faces. To
do this, the slant heights of the
faces, or means of finding them, must be given.
The volume of the wedge is given by the formula:
Fig. 126
MENSURATION OF SOLIDS 259
162. Prismoid — An irregularshaped solid having five or
more flat or plane faces, two of which are
parallel.
Volume of prismoid given by the /,£
formute: V = ^{A + B\U1\,
o
where A and B are the areas of the FlG 127
parallel faces, M the area of a section
halfway between them, and h the height.
Exercises XCVII.
1. Find the weight of a brass wedge, whose height is 5"
and edge 4", the base being a rectangle which measures 8"
by 6" (1 cu. in. = 3 lb.).
2. How many tons of earth are removed in excavating a
trench of which the top and bottom are rectangles? At the
top it is 400' long by 18' wide, and at the bottom it is 350'
long by 15' wide. The bottom is horizontal and the depth
12', (given 1000 cu. ft. earth weighs 40 tons).
3. In a castiron wheel the inner diameter of the rim is
2' and the crosssection of the rim is a circle of 6" radius; find
the weight of the rim.
4. The crosssection of the rim of a castiron flywheel is
a rectangle 8" by 10". If the mean diameter is 10', find the
weight of the rim.
5. A wedgeshaped trench is 40 yards long at the top and
8' wide; the length of the bottom edge is 32 yards and the
depth is 10'. How many cu. yd. of earth have been excavated?
6. The crosssection of the rim of a flywheel is a rectangle
6" by 8", the shorter dimension being in the diameter of the
wheel. The wheel is 22' in outer diameter; find the weight
if the specific gravity of the material is 7 • 2.
Miscellaneous Exercises XCVIII.
(1 cu. in. castiron = 26 lb. 1 cu. in. steel = 2834 lb.)
1. A square bar of wroughtiron 12' long, weighs 80 lb.
What is the size of the end?
2. A closed castiron tank is 3' long, 2f wide, and 2§'
deep, outside measurements. If the material is \" thick,
find the weight of the tank.
260 MATHEMATICS FOR TECHNICAL SCHOOLS
3. The rain which falls on a roof 22' by 36' is conducted to
a cylindrical cistern 8' in diameter. How great a rainfall
would it take to fill the cistern to a depth of 7'?
4. Water is poured into a cylindrical reservoir 20' in
diameter, at the rate of 300 gallons per minute. Find the
rate, in feet per minute, at which the water rises in the
reservoir.
5. The internal diameter of a cylinder, open at the top, is
1', and its weight is 180 lb.; when filled with water it weighs
2000 lb. ; find the depth of the cylinder.
6. Find the weight of a copper tube f " outside diameter,
•05" thick, and 5' 10" long.
7. A steel bar whose crosssection is a regular hexagon 1"
to the side, is 8' in length. Find its weight.
8. A trough whose crosssection is an equilateral triangle
8" to the side contains 30 gallons of water; how long is it?
9. A boiler has 275 tubes, each 19' 3" long and 2f" in
diameter. What is the total heating surface of the tubes?
10. Find the capacity in gallons of a conical vessel 15" in
diameter and 2' in slant height.
11. A conical tent covers an area of 154 sq. ft. and is 6'
in height. How many sq. yd. of canvas does it contain?
12. What is the volume of a cylindrical ring having an
outside diameter of 6", an inside diameter of 5 A", and a
height of 5f "?
13. Water flows at the rate of 20' per min. from a cylindri
cal pipe 25" in diameter. How long would it take to fill a
conical vessel, whose diameter at the surface is 10" and depth
9"?
14. The external diameter of a hollow steel shaft is 20",
and the internal diameter 12". Find the weight of 20' of
this shafting.
15. From a cylinder whose height is 8", and diameter 12",
a conical cavity of the same height and base is hollowed out.
Find the whole surface of the remaining solid.
16. Find the cost of polishing the lateral surface of a
pyramid 6' 5" high, standing on a square base 6' to the side,
at the rate of 20 c. a sq. ft.
17. How many gallons of water will be discharged per
min. from a 4" pipe if it flows at the rate of 300' per minute?
MENSURATION OF SOLIDS 261
18. The crosssection of a water pipe is a regular hexagon
whose side is 1". At what rate, in feet per min., must the
water flow through the pipe in order to fill in one hour a
cylindrical tank the radius of whose base is 16'' and whose
depth is 5'?
19. The base of a prism whose altitude is 15" is a quadri
lateral whose sides are 10", 18", 12", 16", the last two forming
a rt. angle. Find its volume.
20. A tower whose ground plan is a square on a side of
30', is furnished with a pyramidal roof 8' high. Find the
cost of covering the roof with sheetiron at 25c. a sq. ft.
21. A steel bar whose crosssection is an equilateral triangle
1\" to the side is 8' long; find its weight.
22. A cylindrical granite pillar 10' high and 30" in diameter,
is surmounted by a cone 2\' high. Find the weight of the
whole if a cu. ft. of granite weighs 165 lb.
23. How many cu. in. are there in a hexagonal blank
nut 5" to a side and f" thick?
24. It is desired to make a conical oil can with a base 5"
in diameter to contain \ pint; what must be the height?
25. A piece of castiron has a B. & S. taper — \" to 1'. It
is 10" long and the diameter at the large end is 35"; find its
weight.
26. Find the height of a pyramid, of which the volume is
625 cu. in. and the base a regular hexagon 12" to the side.
27. The perpendicular height of a square chimney is
150' 3". The side of the base measures 12' 6" and the side
at the top 6' 3", the cavity is a square prism whose side meas
ures 3' 9". How many cu. ft. of masonry in the chimney?
28. A circular disc of lead, 3" in thickness and 12" diameter,
is converted into shot, each 05" in radius. How many shot
does it make?
29. The interior of a building, in the form of a cylinder
of 15' 0" radius and 10' 0" high, is surmounted by a cone whose
vertical angle is a rt. angle. Find the area of the surface
and the cubical contents of the building.
30. A square building 20' 0" to the side has a hip roof in
the form of a pyramid. The peak of the roof is 10' above
the plate level and the rafter heel is 2'; find the cost of roofing
with shingles, laid 4" to the weather, material and labour
costing $12 a square of shingles.
262 MATHEMATICS FOR TECHNICAL SCHOOLS
31. The base of a cone is an ellipse, major axis 4", minor
axis 2", height 6". Find the volume.
32. A quart measure is 8" in height. Find the diameter
of its base.
33. Find the weight of a log 40' long, 4' 6" in diameter at
one end and 30" in diameter at the other, the specific gravity
of the wood being 78.
34. A piece of copper 6" long, 2" wide, and \" thick, is
drawn out into a wire of uniform thickness and 100' long.
Find the diameter of the wire in mils.
35. A conical vessel 7" deep and 20" across the top is
completely filled with water. If sufficient water is now
drawn off to lower the surface 6", find the area of the surface
of the vessel thus exposed.
36. A cylinder 2" in diameter and 8" in height contains
equal volumes of mercury, oil and water. If the specific
gravity of the mercury be 136, of oil »92, find the total
weight of contents.
37. The radii of the internal and external surfaces of a
hollow spherical shell of metal are 10" and 12" respectively.
If it is melted down and the material formed into a cube,
find the edge of the cube.
38. An automobile gasoline tank has an elliptical cross
section 9" by 15" and is 3' long. How many gallons of gasoline
will it hold?
39. A hemispherical basin holds 2 gallons. Find its
internal diameter.
40. If 30 cu. in. of gunpowder weigh 1 lb., find the internal
diameter of a spherical shell that holds 154 lb.
CHAPTER XVII.
RESOLUTION INTO FACTORS.
163. When a quantity is the product of two or more quantities,
each of these is called a factor of the quantity, and the finding
of these quantities is called factoring the quantity.
Thus, 6 is the product of 3 and 2, therefore 3 and 2 are called
the factors of 6.
Further 2xy is the product of 2, x and y, therefore 2, x and y
are called the factors of 2xy.
164. First Type. If we wish to find the value of 6X4+6 X3,
we could find the value of 6X4 = 24, then 6X3 = 18, and then
add the results giving 42.
We might also write as follows:— 6X4+6X3 = 6(4+3) =
6X7=42.
The following examples will illustrate this algebraically:
1. bx\by\bz = b(x{y\z). 3. 4a; 2 — 16x?/ = 4.r(x — 4^).
2. 2^+4 = 2(?/+2). 4. a6+ac+a = a(6+c + l).
Frequently an expression may be resolved into factors under
this type by arranging in groups which have a compound
factor common.
Thus, factor x 2 — ax+bx — ab.
By taking x out of the first and second and b out of the third
and fourth we may write as follows: — x {x — a) +6 (x — a).
We have now the sum of two products with x — a in each,
therefore we may write as {x — a)(x+6).
Examples:
1. Factor a 2 +ab+ac+bc. 2. Factor 12a 2 4a63ax 2 +6x 2 .
= a(a+6)+c(a+6). =3a (4ax 2 ) &(4az 2 ).
= (a+b)(a+c). =(4az 2 )(3a&).
263
264
MATHEMATICS FOR TECHNICAL SCHOOLS
Factor:
1. ax — a 2 .
2. x 2 3ax.
3. 5x 3 15x 2 y.
4. 8a 3 16a&.
5. 2156x.
6. ay + by+cy.
7. ax — bx — ex.
8. 3a 2 6 2 9a6 + 12.
9. 14x 3 7x 2 y456x?/ 2
10. 5a 2 + 15ax+20a&.
Exercises XCIX.
11. ax — bx — ay\by.
12. x 2 — xy+xz — yz.
13. 3x — 3y{ax — ay.
14. x 3 x?/2x 2 42?/.
15. ab (x 2 + l)x (a 2 +6 2 ).
16. a 5 + a 4 + a + l.
17. a 2 bcb + a 2 c.
18. 2a 3 +6a 2 ca3c.
19. x 2 + mx (m + l) + m 3 .
20. ax+6x+a«/+6?/ — az — bz.
165. Second Type. In the treatment of multiplication we
found the product of two binomials asx+2andxf5 as follows:
x+2
x+5
x 2 +2x
+5x + 10
a; 2 + 7x + 10*
While the result could always be obtained by this method,
it is important that the student should be able to write down
the product of two binomials by inspection. In the result
above we observe that the first term is the product of the first
terms of the two expressions; the third term is the product
of the second terms of the two expressions; the middle term
has for its coefficient the sum of the numerical quantities
(with proper sign) in the second terms of the two expressions.
Write down the values of the following products:
1.
0r+4)(:r45).
7.
(x — 3a)(x + 2a).
2.
06)(xf2).
8.
(x+7ij)(xSy).
3.
(p+3)(p6).
9.
(2x5)(2x + 6).
4.
(r+4)(r6).
10.
(3.rl) (3x41).
5.
(x+6)(x+3).
11.
(2x+7y)(2x5y)
G.
(p9)(p + l).
12.
(2x4a)(2x+6).
RESOLUTION INTO FACTORS 265
The converse problem gives us our second type and consists
in finding the two factors if we know the product.
Thus, factor z 2 +7z + 12.
The second terms of the factors must be such that their
product is + 12 and their sum +7. Hence they must both be
positive, and it is readily seen that they must be +4 and +3.
.. z 2 +7x + 12 = (x+4)(x+3).
Factor x 2 —10ax + 9a 2 . •
The second terms of the factors must be such that their
product is 9a 2 and their sum — 10a. Hence they must be
— 9a and — a.
.. z 2 10a 2 +9a 2 = (x9a)(za).
If we multiply 3x+4 by 2x + l we get 3x(2x + l)+4(2a; + l)
= 6x 2 +3x+8z+4 = 6z 2 +ll:r+4.
The converse problem is now to be considered.
Factor 4x 2 +llz 3.
Here the numerical coefficients of the first terms of the
factors must be 4 and 1, or 2 and 2, and the last terms must
be 3 and 1.
The possible sets (omitting the signs) are:
4x 3, x 3, 2x 1.
x 1, 4a; 1, 2x 3.
Since the sign of the last term in 4x 2 +llx — 3 is minus,
we at once decide that the signs of the last terms in the factors
must be different, and therefore that the partial products
must be subtracted. The second arrangement is the only
one from which we can obtain llx, and also since the middle
term is positive, the larger of the cross products must be
positive.
.*. 4z 2 +llz3 = (x+3)(4xl).
Example 1:
12x 2 x20=(3x4)(4a;+5).
Example 2:
Factor 3a; 2 7x+2 = (3xl)(a;2).
266
MATHEMATICS FOR TECHNICAL SCHOOLS
Factor and verify:
1. z 2 +10x + 21.
2. x 2  IQx + 24.
3. x 2 4x+4.
4. x 2 x2.
5. z 2 llx + 10.
6. z 2 a;42.
7. x 2 3x 130.
8. l3x+2x 2 .
9. z 2 +x72.
10. z 2 +4x5.
11. 5 — 4x — x 2
12. 4013:r+a; 2 .
13. l5x+6z 2 .
14. 403zx 2 .
15. l3x130x 2 .
Exercises C.
16. 5x 2 +42z27.
17. 4z 2  16z + 15.
18. 3x 2 22x+7.
19. 6x 2 llx+3.
20. 9z 2 9x28.
21. 26x 2 4Lr+3.
22. 12.r 2 17xf5.
23. 5x 4 10x 2 ?/ 2 400?/ 4 .
24. 2x 2 +5xy+Sy 2 .
25. 12x 2 2^30.v 2 .
26. 12x 2 5:n/3?/ 2 .
27. 8x 2 +22x+9.
28. 6x 2 l3xy+6y 2 .
29. 13xV9x 4 4?/ 4 .
30. Ux 2 +8Sxy6y 2 .
166. Third Type.
If we multiply x+y by x— y we have:
z 2 +:n/
— xy — y 2
x 2 —y 2
Observing the above we find that, when we multiply the sum
of x and y by the difference
of x and y, the result is the
difference of the squares of
x and y. Therefore we may
say that the difference of
the squares of two quantities
is equal to the sum of the
quantities multiplied by the
difference of the quantities.
h« a
* « —
C
n
t
,
M
)
i
Fig. 128
Geometrical Illustration:
(a+6) (ab) = ANHC
= BMHC+CKFD
= BEFDMEKII
= a 2 6 2 .
RESOLUTION INTO FACTORS 267
Examples:
1. p 2 q 2 =(p+q)(pq).
2. 9a 2 256 2 = (3a) 2 (56) 2 =(3a+56)(3a56).
3. TR 2 irr 2 =T(R 2 r 2 ) = ir(R+r)(Rr).
4. (a+6) 2 c 2 = (a+6+c)(a+6c).
167. Incomplete Squares. Sometimes an expression comes
under this type, but it is not stated directly as the difference
of two squares.
Thus, factor a 4 +a 2 6 2 +6 4 .
This expression would be the square of a 2 f6 2 if the middle
term were 2a 2 6 2 instead of a 2 b 2 .
We will then add a 2 b 2 to complete the square and subtract
it again to maintain the value of the expression.
Thus, a 4 +a 2 6 2 +6 4 = (a 4 + 2a 2 6 2 +6 4 )a 2 6 2 =(a 2 +6 2 ) 2 (a6) 2 .
= (a 2 + b 2 +ab)(a 2 +b 2 ab).
Examples:
1. z 4 +x 2 + l = {x i +2x 2 + l)x 2 .
= 2 + l) 2 x 2 .
= {x 2 + l+x){x 2 + lx).
2. z 4 +9x 2 +25 = Or 4 +10z 2 + 25)a; 2 .
= 2 +5) 2 z 2 .
= (x 2 +5+x)(x 2 +5x).
3. 4x 4 + l = (4:r 4 +4.r 2 + l)4:r 2 .
= (2z 2 + l) 2 (2z) 2 .
= (2x 2 + 1 + 2x) (2x 2 + 1  2x) .
Exercises CI.
Factor:
1. lQx 2 25y 2 . 6. x i y i .
2. x 2 9y 2 . 7. x s y s .
3. 25zy16a 2 6 2 . 8. a 2 6 2 26cc 2 .
4. 9x 2 y 2 4:p 2 q 2 . 9. (x+y) 2 {a+b) 2 .
5. a 2 (6+c) 2 . 10. (x 2 +y 2 ) 2 ±x 2 y\
268 MATHEMATICS FOR TECHNICAL SCHOOLS
11. x 2 y 2 +2yzz 2 . 20. x 4 +9x 2 + 81.
12. la 2 2abb 2 . 21. x 4 +4?/ 4 .
13. a 16 l. 22. x*7x 4 + l.
14. x 2 2:n/ + ?/ 2 a 2 2a&& 2 . 23. 4.r 4 37xV +9*/ 4 .
15. 7rl0 2 7r7 2 . 24. x 4 +4x 2 +16.
16. 7r75 2 7r25 2 . 25. 9x 4 IOxV+i/ 4 .
17. 5a 2 10a6+56 2 20c 2 . 26. 4a: 4 13x 2 2/ 2 +92/ 4 .
18. 16a 2 6 2 +2a6. 27. x 4 +5x 2 2/ 2 +9?/ 4 .
19. 4x 4 + llxV+9?/ 4 . 28. x 4 +x 2 f25.
168. Fourth Type. Divide x 3 \y 3 by x+y.
x+y)x 3J ry z /x 2 — xy+y 2
x 3
x
x 3 \x 2 y
— x 2 y+y 3
— x 2 y — xy 2
Divide
xy 2 +y 3
xy 2 +y 3
y 3 by x — y.
y)x 3 —y 3 /x 2 +xy+y 2
x 3 — x 2 y
x 2 y — y 3
x 2 y — xy 2
xy 2 — y 3
xy 2 y 3
As a result of the above we may write :
x 3 \y 3 = {x\y)(x 2 — xy\y 2 ) and x 3 — y 3 = {x—y){x 2 — xy\y 2 ).
The above results might be stated as follows:
The sum of the cubes of two quantities is divisible by the sum
of the quantities, and the difference of the cubes of two quantities
divisible by the difference of the quantities. The other factor
consists of the sum of the squares of the quantities, minus their
product, if the sum of two cubes, and plus their product if the
difference of two cubes.
RESOLUTION INTO FACTORS 2(59
Examples:
1. p 3 +q z = (p+q)(p 2 pq+q 2 ).
2. 8x 3 27y 3 =(2xy(Zyy = (2x3y)(4:X 2 +Qxy+9y 2 ).
3. 5a 3 40 = 5(a 3 8) = 5(a2)(a 2 +2a+4).
Exercises CII.
Factor:
1. 2/ 3 +27. 5. x 6 64. 9. 2x 3 f250.
2. a 3 125. 6. a 3 216. 10. x l2 y 12 .
3. x 6 + l. 7. 381x 3 . 11. (a+6) 3 c 3 .
4. a 6 6 6 . 8. z 4 27z. 12. (a+6) 3 (o6) 3 .
CHAPTER XVIII.
INDICES AND SURDS.
169. Indices. In the introductory chapter in Algebra we
inferred the laws with respect to indices from particular cases.
Thus, (1) x 3 Xx 2 = x 3+2 = x 5 . (3) (f) 2 = x 3 Xx 3 =i 8 .
(2) x 5 Hx 2 = x 52 = x 3 . (4) (xy) 2 = xyXxy = x 2 y 2 .
In the following discussion general proofs will be given for
these laws and also their application when the indices are
fractional, zero, or negative.
Definition. If x is any number and m any positive integer
x m means the product of m factors each equal to x.
1 . To prove x m X x n = x m+n 
By definition:
x m Xx n = (xxxxx. ...torn factors) X (xxxxx to n factors).
= xxxxx to (rafn) factors.
= x m+n ky definition.
From the above it follows that:
x m xx n xx v = x mJrn xx v = X m ^~ n ^~ V m
x m
2. To prove — =x m ~ n m>n.
By definition:
x m
— = (xxxxx. ...torn factors) j (xxxxx ton factors).
= xxxxx to (ra — n) factors.
= x m ~ n by definition.
3. To prove {x m ) n = x mn 
{x m ) n = (x m ) X (x m ) X (x m ) to n factors.
_ gjn+m+m .... to n terms k„ i
= x mn 
270
INDICES AND SURDS 271
4. To prove (xy) m = x m y m 
(xy) m = (xy) X (xy) X (xy) .... to m factors.
= (xxxxx. . to m factors) X (yxyxy. ..torn factors).
= x m Xy m = x m y m .
The above are known as the fundamental laws of indices.
In assigning a value to x m , the definition requires that m
be a positive integer, so that x*, x~ 3 , x° have as yet no meaning.
However, one of the advantages of Algebra over Arithmetic
is that it extends the principles of Arithmetic to negative
numbers, so in harmony with this principle we will assume
that the laws proved for positive integral indices holds for
negative and fractional indices.
(a) Meaning of x°.
Since x m Xx n = x m+n for all values of m and n, if we replace m
by 0, we have x° Xx n = x n+0 =x n .
x n
x n
This relation was assumed when we said that log 1=0,
for by the above 10 = 1, therefore by definition of logarithm,
log 1=0.
(6) Meaning of x~ n .
Since x m Xx n = x m+n for all values of m and n, if we replace
m by — n, we have:
x~ n Xx n = x~ n+n = x° .
But x° = l.
x n
From the above it follows that any factor may be transferred
from the numerator to the denominator of an expression, or vice
versa, by changing the sign of the index.
272
MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises CIII.
Express with positive indices:
1. x~ 2 . 4^
2. p~ e . 2*'
1 „ ■ 5"
a; 3 *
2
3  2 
6 1
3 2 
3.
4.
5.
10 ^xS^A
25 l *
a~ 4 "
11. 3~ 2 xjx3 3
1
12
xa 3 x— .
1 2 „a*
9. a 6 x— x— 
a z a
(c) Meaning of x q , p and q being positive integers.
Since x m xx n = x m+n for all values of m and n, if we replace
both m and n by §, we have x i xx i = a; J+i = a; 1 = a\
Thus if x^ be multiplied by a;* we get the product x, or
otherwise stated the square of x* = x.
We have, however, previously represented the quantity
whose square is x by y/x.
.'. X i = y/x.
Similarly x^xx^xx* = # i+i+ * = x,
.'. x* = \/x (cube root of x).
Generally xn = Va;(nth root of x).
Again, since (x m ) n =x mn ,
then (x*) 4 =x 3 ,
.*. x$ = \/x 3 .
Example 1
Example 2
Example 3
Similarly, (x«) q = x p , ,
.*. x"= 1/x p , p and q being positive integers.
16* = ^16 = ^2* = 2.
27* = (^27) 2 = 3 2 = 9.
64* = (a$/64) 5 = 2 5 = 32.
INDICES AND SURDS 273
Exercises CIV.
Write with positive indices:
1. a 3 6~ 2 .
6. 2x*X3aT 1 .
ii. ,/—
12. 2.
4r _1
13. _, .
x *
14. 7a*X3a" 1 .
9 a " 2 v a3
*' b**b*'
a 3
3.  2 X a~*.
or
a" 2 6 8
*' c^6*
7. 2ai
8.4,.
x J
« 2a 2
9. r.
a  *
_ 2x~ l
5 ' 4jT»"
io. JL.
Vz 3
a  *
6a
Ifa = l, 6 =
2, n =
= 3, find the value
of:
16. (ab) n .
19. (a n b n )\
22. (a 3 6 3 ) n .
17. <£)\
20. (a'ft 1 )  ".
23. (a" 4 6) n .
18. (a*by.
21. (a" 2 6 2 ) n .
Find the value of:
24 2X6 " 2
J4 ' 3" 2 '
27 ' F*
30. (A 2 *)" 1 .
25. 16?.
28  25"*
28. (If)"*.
31. (W) 1 .
29. 16 1 " 5 .
32. 36"i
Show that:
33. 12* = 2X3
i
34. 108 4 = 3X2'
I
35. 80* = 2X5*.
Express as
the root of an integer:
36. &XS*.
37. 3 s X 9*.
38. 3* X 9* 4 27*.
39. Multiply
x>+y h hy x h y h . 40.
Multiply z* +2/* by x — y.
Solve:
41. x h = 2.
42.
aT* = 4. 43.
1
= 4. 44. z* = 27.
170. Surds.
Definition. If the root of a number cannot be exactly
determined, the root is called a surd.
274
MATHEMATICS FOR TECHNICAL SCHOOLS
Fig. 129
Thus, y/2 is a surd because we cannot find a number whose
square is exactly equal to 2.
We can find its value to
a number of decimal places
(14142), but this is only an
approximate value.
We can, by a geometrical
process (Fig. 129), find a line
which is the y/2 units in length.
If we draw two lines at right
angles to each other and each
1 unit in length, then the
hypotenuse of the rightangled
triangle so formed would be
the y/2 units in length. By
continuing as in diagram, lines a/3, \/4, etc., may be found.
171. Quadratic Surds. We are chiefly concerned with surds
in which the square root is to be found. These are called
t
quadratic surds.
Thus, \/2, a/3, V6, a/8 are quadratic surds.
172. Surds other than Quadratic. These are indicated by
the root symbol.
Thus, ^6, ^9, AyiO, the first being called a surd of the
third order, the second a surd of the fourth order, the third a
surd of the fifth order.
A surd is sometimes called an irrational quantity, and for
the sake of distinction, quantities which are not surds, are
called rational quantities.
173. Like and Unlike Surds. When surds in their simplest
form have the same surd factor they are called like surds,
otherwise they are unlike surds.
INDICES AND SURDS 275
Thus, 2a/3~, 3V3", 5 a/3 are like surds, and 2 a/3, 3a/2^
5 a/6 are unlike surds. Just as we add and subtract like terms
in Algebra, so we may add and subtract like surds.
Thus, 2 a/3 +3 a/3 = 5 V3.
5a/23a/2 = 2a/2. 5 a/6 +2 a/6 3 a/6 = 4 a/6.
174. Multiplication of Surds. Since a/3 represents a quantity
whose square is 3, .*. a/3 X a/3 =3.
Again, since (a/3X a/2) 2 = a/3X a/3X a/2 X a/2.
= 3X2 = 6.
.'. a/3 X a/2 = a/6.
Similarly, a/3 X a/5 = V lT.
Generally, a/« X a/& — \/a&
In the above the surds multiplied together are of the same
order. If, however, we wished to multiply a/2 and a/3, it
would first be necessary to change them to surds of the same
order.
By the previous section on indices:
V2 = 2* = 2?, a/3 = 3* = 3*.
.*. V2X^3 = 22X3* = Ay2 3 XAy3 2 .
= a/8 X a/9 = a/72.
Exercises CV.
Express as surds of the same lowest order:
1. a/3, a/4, a/6.^ 3. a/2,V8, a/4.
2. a/5\ \/ll, a/18, 4. a/* a/3", V6.
Find the product of:
5. a/2, a/3. 8. a/5, a/6.
6. a/3, a/5, a/2. 9. a/2, a/3, a/4.
7. Vf, Vf, Vf 10. V3, V5, a/6.
276 MATHEMATICS FOR TECHNICAL SCHOOLS
175. Mixed and Entire Surds. When a surd quantity is
the product of a rational quantity and a surd, it is called
a mixed surd. If there is no rational factor it is called an
entire surd.
Thus, 6\/3 is a mixed surd, and \/7 is an entire surd.
The expressing of a mixed surd as an entire surd would
be of little value practically, but the reverse process is of
frequent application.
Thus, V27 = V9X3=3 v / 3.
Again, V72= V36X2 = 6\/2.
Exercises CVI.
Express as a single surd:
1. 2 V63 + 5 V28  V7. 3. V72 + V98  V128 + V32 + V50
2. 10V444V99. 4. V45V20 + V80.
Find the value correct to two places of decimals:
5. V288. 11. V36V72 + V90.
6. V147 12. 4V63+5V78V28.
7. V250. 13. 2V3635v / 2434V'192.
8. 3V150. 14. 5V242V54V6.
9. 5V245. 15. 4V128+4V755V162.
10. 4V63".
Express in simplest form:
16. ^256. 17. ^432. 18. ^3125. 19. ^/2187.
Find the value to two decimal places:
20. 2V14XV2T. 24. 2Vl4X3\/28.
21. 3V8XV128. 25. 2V15X3V5.
22. V50XV75. 26. 8V12X3V2T
23. 3V6X4V2.
INDICES AND SURDS 277
176. Division of Surds.
Since y/x X Vy = Vxy,
Si mil arly , y/x i y/y = % ,
\y
and 2V3043V6 = 3\'y = ^V5.
Example: — Find the numerical value of— j= (tan 30°).
We might find the square root of 3 and perform the division.
This, however, would not be the best method,
ForUJ_ x V3 = V3 = L7321 = . 5774 .
V3 \/3 V3 3 _3
1 V3
Here we changed ~~k into J ~ by multiplying both numera
v o o
tor and denominator by \/3.
This operation of making the denominator a rational
quantity is called rationalizing the denominator.
1
Example: — Find the value of — ;= (sin 45°).
UJLx^^i^Ton.
V2 V2 x/2 2 2
Example: — To rationalize the denominator of an expression
of the form ^^
2V2.
Here we wish to convert = into an equivalent expres
2 — s/2
sion but with a rational denominator.
Since the product of the sum and difference of two quanti
ties is equal to the difference of their squares, then (2 — \/2)
(2 + V2) =42 = 2.
. l + V2 x 2 + v / 2 _ (l + V2)(2 + V2) ^ 4+3V2
2V2 2 + V2 2 2
278 MATHEMATICS FOR TECHNICAL SCHOOLS
The expression 2 + \/2 is known as the conjugate expression
to 2 — \/2. If the denominator of the fraction had been 2 + y/2
we would then have multiplied by 2— V2.
Exercises CVII.
Calculate the value of the following to 3 places of decimals:
1. *L 8. U 15. *****
V3 V500 7V23
2 4 2
2 V3* 9 'VW 16  3 "v6
„ 12V2 ,n /25 
3 ^' 10 'V252 17. (V3V2) 2 
4 A 11. i^ 18 . V31
V5 2 "V2 V21
5 4= 12 =5— • 19 3 ^1 1
V24* V5 + V2 " 3V21*
48 13 5+2V6 . 4V7+3V2
V6^_ 62V6 V3V2
256 14 j
1575* ' V51*
V
CHAPTER XIX.
QUADRATIC EQUATIONS.
177. Quadratic Equations. The following problems will lead
to equations which differ somewhat from those previously
solved.
Problem 1:
The area of a square is 64 sq. in. What is the length of
a side?
If x represents a side of the square, then x 2 = 64.
In the equations previously met the unknown x occurred
to only one power, and that the first. Here, however, the
unknown occurs to the second power. When an equation
contains the square of the unknown quantity, but no higher
power, it is called a quadratic equation.
In the equation x 2 = 64 we have the simplest form of the
quadratic equation:
If x 2 = 64.
then x = ±S.
We have here two values of x, i.e., + 8 and— 8, which will
satisfy the equation. If we regard f and — as opposite
directions in the same straight line, the minus value has no
significance in determining the side of the square.
Problem 2:
The length of a numberplate on a machine is 6" more than
its width. If its area is 72 sq. in., find its dimensions.
If x = No. of in. in width,
then x + 6 = No. of in. in length,
then x (x + 6) =72.
or x 2 + 6.r72 = 0.
279
280 MATHEMATICS FOR TECHNICAL SCHOOLS
By our previous principles in factoring x 2 \6x — 72 = (arf12)
(x6).
Now, if z 2 +6z72 = 0, then (a? + 12)(a:6)=0.
In order that the product of these two factors may be equal
to zero, it is necessary that one factor should be equal to
zero.
Thus the equation will be satisfied if x + 12 = or x — 6 =
or if x = —12 or 6.
As + 6 is the only admissible value, therefore the width
= 6" and the length 6+6 = 12". The equation z 2 +6.r72 =
is known as a complete quadratic equation, containing as it
does both the square and the first power of the unknown
quantity.
Exercises CVIII.
Solve the following equations and verify:
1.
x 2 = 49.
11.
x 2 2 = x.
2.
x 2 +3x = 0.
12.
4x =45 — x 2 .
3.
x 2 = 7x.
13.
5x 2 12z+4 = 0.
4.
z 2 4 = 0.
14.
3z 2 + 14x15 = 0.
5.
6x 2 = 54.
15.
20x 2 +4Lr+20 = 0.
6.
z 2 3 = l.
16.
5+9x2a; 2 = 0.
7.
z 2 10z+21 =
0.
17.
18a: 2 9x 2 = 0.
8.
z 2 14z+48 =
0.
18.
13z 2 +4Lt + 6 = 0.
9.
a; 2_ a ._20 = 0.
19.
12a: 2 x 20 = 0.
10.
z 2 +10 = llx.
20.
6x 2 x2 = 0.
178. Solving by Completing Squares. In connection with
the squaring of a binomial we recall that (a + 6) 2 = a 2 + fe 2 + 2a6,
or that the square of a binomial equals the square of each
term, plus twice their product. If then we have x +6.r and
we wish to add a sufficient quantity to make a complete square,
we could reason as follows: # 2 is the square of x, Qx is twice
the product of x and 3, therefore it is necessary to add 3 2 or 9.
.*. a: 2 +6x+9 is a complete square = (x+3) 2 .
QUADRATIC EQUATIONS 281
Similarly, to x z — 8x or x 2 2XxX4 we must add 4 2 or
16, giving x 2 — 8x + 16 = (x — 4) 2 .
Again, to x 2 +Qx, or x 2 +2X^X we must add (f) 2 or ^,
giving a; 2 +9a;+^ = (a;+f) 2 .
An analysis of the three cases above would lead us to infer
that we completed the square in each case by adding the square
of half the coefficient of x.
This method is necessary where the quadratic equation
cannot readily be resolved into factors.
Thus, Example 1:— Solve x 2  6a;  13 = 0.
or, a: 2 — 6a; =13.
Completing the square on the lefthand side we have:
a; 2 6a;+9 = 13+9.
or, (x3) 2 = 22.
Extracting square root, x — 3 = ±22.
.*. x = 3 + V22 or 3V22.
= 769 or 169.
Example 2:— Solve 3a; 2 +8a; + 12 = 0.
In the examples above on completing the square, we observe
that the coefficient of x 2 in each case is unity and further that
it is positive. Before attempting then to solve this equation
we must make these two changes.
3x 2 +8x + 12 = 0.
= 3a; 2 8a; 12 = 0.
= x 2 — fa;— 4 — 0.
Complete the square, giving:
* 2 t*+a) 2 =4+(!) 2 .
or, (a;!) 2 ^^
or, *$'* ±V¥'<
or, x = $±V^.
= l + VVorV¥.
= 374 or 107.
282 MATHEMATICS FOR TECHNICAL SCHOOLS
Exercises CIX.
Solve by completing the square:
1. 5x 2 + 14a; 55 = 0. _J_ 1 _ 6
2. 9x 2 1436x = 0. l+a;~3^a; = 35'
3. 19x = 158a; 2 . 5 4 3
15.
4. 6x 2 9x15 = 0. x2 x~x+6"
5. 5x 2 +llx12 = 0. _^_ 5 3
6. 2x 2 + 79x = 0. lb ' x l~a;f 2~x"
_*±3_2xl
2xl x3
18. 21x 2 2ax3a 2 = 0.
19. 12x 2 +23fcx + 10fc 2 = 0.
1 5 2
7.
5x 2 15x + ll =
8.
x 2 7x+5 = 0.
9.
x 2 +ll = 7x.
10.
4x 2 = ^x+3.
11.
/v.2 O — 2 3/h
12.
a; — 1
13.
5a; — 1 _ 3x
x + 1 2 '
20.
2x — 5a 2x — a
179. The General Quadratic Equation. From the preceding
examples it is apparent that every quadratic equation can be
reduced to the form
ax 2 f&x+c = 0,
where a, b, c may have any numerical values whatever. If
then we would solve this general quadratic equation we could
use the result as a formula to solve particular cases and con
sequently save the labour entailed.
ax 2 + 6x+c = 0.
Transposing, ax 2 +bx= —c.
dividing by a, x 2 \ — x = — .
Completing the square by adding to each side the square
of half the coefficient of x, i.e., («) '
bx . / b \2 b 2
, . OX ( \2 0' C
giving * 2 + a +( 2a ) = 4a2 " a
or,
(■*=)'
2a> 4a 2
QUADRATIC EQUATIONS 283
Extracting the square root
Hence, x =
b ±Vb 2 4ac
2a 2a
±V& 2 4ac
2a
We might here restate the steps required in solving a quad
ratic of the above form.
(1) Simplify the equation so that the terms in x 2 and x are
on one side of the equation, and the term without x on the other.
(2) Make the coefficient of x 2 unity and positive by dividing
throughout by the coefficient of x 2 .
(3) Add to each side the square of half the coefficient of x
thus completing the square.
(4) Take the square root of each side.
(5) Solve the resulting simple equations.
Example 1: — Solve x 2 — 5x — 3 = 0.
Here, a = l, b= — 5, c= — 3.
5±V(5) 2 4XlX(3)
.*. x = ~ •
= 5±V25 + 12
2
_ 5±V37 _ 5+608
" 2 2
= 554 or— 54.
Example 2 :— Sol ve x 2  2x + 5 = 0.
Here, a = l, 6= —2, c = 5.
2d=V(2)4XlX5
X = 7i •
2±V420
2
2±V16
284 MATHEMATICS FOR TECHNICAL SCHOOLS
In the preceding result the numerical value of the roots
cannot be found, as there is no number whose square is
negative.
Such a quantity as y/ — 16 is called an imaginary quantity,
and the roots are said to be imaginary. This is equivalent
to saying that there is no real number which will satisfy the
equation x 2 — 2x + 5 = 0.
180. There are some Equations that are not really
quadratics but may be solved by the methods of this
chapter.
Example 1: — Solve x 4 — 5x 2 +4 = 0.
Factoring (x 2 — 4) (x 2 — 1) = 0.
.\ x 2 = 4 or x 2 = 1.
.'. x = ±2 and x = ±l.
72
Example 2 :— Solve x 2  x + j—  = 18.
Write y for x 2 — x, then we have:
72
Hh=18.
or, y 2 18y+72 = 0.
Factoring, (y12)(yQ) =0.
giving y = 12 or 6.
.". x 2 — x = 12 or 6.
If x 2 x = 12 then a; 2 x 12 = 0.
then(x4)(x + 3)=0.
giving x = 4, or —3.
If x 2 — x = 6, then x 2 — x — 6 = 0.
then (x3)(x+2)=0.
giving x = 3 or —2.
QUADRATIC EQUATIONS 285
Exercises CX.
Solve the following examples:
1. 3x 2 17x + 10 = 0. 8. x 4 13x 2 +36 = 0.
3. 2 (x 2 + l)5x = 0. y " 15 +
4. 25x 2 7z86 = 0. 20 _ c
5. 7x 2 +32a;15 = 0. 1U ' x ~^ 6X x 2 +3x "
6. (2xl) 2 = 25. 11. (x 2 42) 2 + 198 = 29(x 2 +2).
7. 10z 2 = 13x + 9. 12. a: 6 19a: 3 216 = 0.
13. A rectangular nameplate for a machine is to be \\"
longer than it is wide and to have an area of 10 sq. in. What
will be its dimensions?
14. Three holes are to be drilled so that they will lie at the
three corners of a triangle ABC, right angled at B. The
distance from A to G is to be 10" and the distance from
B to C is to be 2" more than from A to B. Find AB
and BC.
15. The sides AB, BC, CA of a triangle measure 13, 14,
15 respectively. From A a perpendicular AD is drawn to
BC. If BD measures x, express the length of AD in two
ways. Equate the results and find x.
16. The owner of a rectangular lot 15 rods by 5 rods,
wishes to double the size of the lot by increasing the length
and the width by the same amount. What should be the
increase?
17. A straight line is 10" long. Divide it into two parts
so that the rectangle contained by the whole line and one of
the parts is equal to the square on the other part.
18. S = \gt 2 is the law governing a body falling from
rest, s = space, g = acceleration due to gravity (32 ft.),
< = time in seconds. How long will it take a stone to fall
from the top of the City Hall tower, Toronto, if it be 305
ft. high?
19. S = ut\\gt 2 is the law for a falling body when it has
an initial velocity, u representing this initial velocity. If
a stone be thrown with an initial velocity of 8 ft. per sec.
from the top of the Eiffel tower, 984 ft. high, in what time
will it reach the ground?
28G MATHEMATICS FOR TECHNICAL SCHOOLS
181. Simultaneous Quadratic Equations. The following
problems will lead to simultaneous equations where one at least
is of higher degree than the first.
Problem :
The perimeter of a rectangle is 18'', and its area is 20 sq.
in.; find its length and breadth.
If x represent the length and y the breadth, then:
2z+2y = 18.
or, x + y = 9 (a).
also, xy = 20 (6).
Solution — 1st method.
from (a), y = 9— x.
Substitute in (6), x(9x)=20.
or, 9zz 2 = 20.
or, z 2 9x+20 = 0.
or, (x — 5)(x — 4) =0.
x = 5 or 4.
20 . 20 _
.'. 2/ = =4or T = 5.
x = 5 or 4.
2/ = 4 or 5.
Solution — 2nd method.
(a) 2 = x 2 +2xy+y 2 = 81.
4x(6) = 4xy = 80.
Subtracting, x 2 — 2xy+y 2 = 1.
.*. {xy) 2 =\.
or, x— y = ±l.
x+y=9 x+y=9
x—y = l x—y=—l
2x =10, x = 5. 2x =8, z = 4.
2y = 8, y = 4. 2y =10, x = 5.
.'. x = 5 or 4, ?/ = 4 or 5.
QUADRATIC EQUATIONS 287
Exercises CXI.
Solve the following equations:
1. x+y = 28,
4. x+y = 84,
7.
x 2 +i/ 2 = 178,
a;?/ = 187.
xy = 92S.
a; +2/ = 16.
2. xy = b,
5. x 2 +y 2 = n,
8.
x 2 +y 2 = 185,
xy = \2Q.
xy = 35.
x y =3.
3. xy = 8,
xy = 51S.
6. x 2 +2/ 2 = 89,
xy = 40.
9.
M3,
10. Divide a straight line 7" long into two parts so that the
rectangle contained by the parts may be equal to 12 sq. in.
11. Divide a straight line 12" long into two parts so that
the sum of the squares on the parts may be equal to 74
sq. in.
12. The hypotenuse of a rightangled triangle is 25", and
the perimeter is 56", find the sides.
CHAPTER XX.
VARIATION.
182. Quantities are often related to each other in such a
way that any change in one quantity produces a corres
ponding change in the other.
For example, consider a train travelling with a uniform
speed. If in one hour the train travels 30 miles, then in
two hours it will travel 60 miles, and so on.
We may state this by saying that the distance travelled is
proportional to the time, or that the distance varies directly
as the time.
If we represent distance by d and time by t, the relation
may be expressed by d varies as t.
The symbol oc represents "varies as", therefore we have
d oc t.
If we let d 1} d 2 , d 3 be successive values of d and
h» t*> 'a • • • corresponding successive values of t, then,
we have:
d t , d.
t= or d= T L t.
d i tx <i
also,
d t , d„ A
j = — or d =  ? t.
u 2 *2 *2
also,
d t , d„ J
j= — or d=—*t, etc.
a 3 t 3 t 3
AAA
Let us consider the expressions y, ~t / in the light
t t t 2 t 3
of the above illustration.
288
VARIATION 289
If d, = 60, < x ~2, then^i =^° = 30.
Ifd 2 =90, t 2 =S, then^ 2 =^ = 30.
If d 3 = 120, * 3 =4, then ^=^ = 30.
From the above we see that the ratios — L , ^, ^ are each
t 1 t 2 t 3
equal to 30, therefore we infer that doct becomes d = 30t
or generally d = Constant x t or d = kt (k being a constant).
Example 1:
The circumference of a circle varies directly as its diameter.
A circle 7" in diameter has a circumference of 22", find the
circumference of a circle of 64" diameter.
From the above C = kd,
then, 22 = £7,
or, fc = y,
,.c=fz).
Substituting for D the value 64,
22
C = ^X64 = 20M4".
In the above we observe that the first set of conditions
enable us to find the constant k. The equation is then one
between C and D, and from any value of one of these we can
find the other.
Example 2:
The areas of circles vary directly as the squares of their
radii. If a circle with a radius of 7" has an area of 154 sq.
in., find the radius of a circle with an area of 1386 sq. in.
From the above Acer 2 . .'. A = kr 2 , then 154 = A; 7 2 , giving
kJ*
* 7*
290 MATHEMATICS FOR TECHNICAL SCHOOLS
Then substituting for A the value 1386 we have:
22
1386= r 2 .
.\ r = 21.
Example 3:
The volume of a gas varies inversely as the height of the
mercury in the barometer. If the volume is 22 cu. in. when
the barometer registers 30", what is the volume when the
barometer registers 32"?
Here we have a case of varying "inversely." This means
that an increase in one quantity gives a proportionate decrease
in the other. Hence, when one quantity varies inversely as
another it varies as the reciprocal of the other.
In the above V oc^j or V=kjj.
£1 tl
k
From the conditions given, 22 = — , giving A = 660,
Ox)
then, F= o ° o u = 206 cu. in.
Exercises CXII.
1. The strength of a beam varies directly as the square of
its thickness.
A beam of given length and width and 6" thick carries a
maximum load of 5 tons. What load will a beam of the
same width and length, but 12" in thickness carry?
2. The weight of a substance varies directly as its volume.
A steel bar containing 100 cu. in. weighs 283 lb. What is
the weight of a bar of the same material containing 642
cu. in.?
3. The velocity of a falling body varies directly as the time
during which it is falling. When a body falls from rest, its
velocity at the end of 1 sec. is approximately 32 ft. per second.
Compute its velocity at the end of 15 seconds.
4. The velocity of the rim of a pulley varies directly as its
diameter. A 12" pulley has a rim velocity, at a certain
moment, of 160' per min. What is the rim velocity, at the
same moment, of a 9£" pulley which is keyed to the same shaft?
VARIATION 291
5. The deflection of a beam under a given load varies
inversely as the square of the thickness. If a given beam
carrying a certain load is 5" in thickness and has a deflection
(due to the load) of 2", what deflection will be produced by
the same load if the thickness be 7§"?
6. The weight of a body varies inversely as the square of
its distance from the centre of the earth. If a substance
weighs 10 lb. at sea level (3960 miles from the centre), compute
its weight on the top of a mountain 29,000 ft. above sea level.
7. The areas of circles are to one another as the squares of
their diameters. If a circle with a diameter of 14" has an
area of 154 sq. in., find the diameter of a circle with an area
of 320 sq. in.
8. The pressure per sq. in. on a hydraulic ram varies inver
sely as the square of the diameter, if the total load on the ram
is constant. If a load supported by a hydraulic ram of 8"
diameter gives a pressure per sq. in. of 40 lb., find the pressure
per sq. in. on a 3" ram which supports an equal load.
9. The horsepower of the engines of a ship varies directly
as the cube of the speed. If the horsepower is 1800 at a
speed of 10 knots, what is the power when the speed is 235
knots?
10. The volumes of spheres vary directly as the cubes of
their diameters. If a sphere with a diameter of 6" has a
volume of 1134 cu  m > find the diameter of a sphere whose
volume is 616 cu. in.
11. The number of rivets required for a boiler seam varies
inversely as the pitch (the distance between rivet centres).
If 35 rivets are required when the pitch is 2 ", determine the
pitch when 40 rivets are required for a boiler of the same size.
12. The resistance of a wire varies inversely as the square
of its diameter.
The resistance of a coil of copper wire \" in diameter was
3 ohms. What is the diameter of a wire of the same length
with a resistance of 23 ohms?
183. Problems involving more than Two Variables. If we
take two rectangles of the same width, it is readily seen
that their areas vary as their lengths. If again we take two
rectangles of different widths but the same length, it is
further agreed that their areas vary as their widths.
292
MATHEMATICS FOR TECHNICAL SCHOOLS
We now wish to consider the variation in the area when
both the length and width vary.
i
I— I
A,
W
Fig. 130
In Figure 130 above the rectangles (1) and (2) have the
same width but different lengths, while the rectangles (2)
and (3) have the same length but different widths.
If A, A 1} and A 2 represent the respective areas, then with
the above notation:
A Iw _ I
A 1 ~l 1 w l t
A x l x w w
A 2 l 1 w l w x
also,
(a),
, '\ w/1 v A A. I w
(«)X(6) jxjij*.
A Iw . A o ,
or,  . = : ■ or A = , — — Iw.
A 2 l 1 w x l 1 w l
Since, A a =l 1 w l .'. j—* =1 (constant).
l l w l
.'. A = Constant X Iw.
.". A oc Iw.
From this we state that the area of a rectangle varies as
the product of its length and width, when both the length
and width vary.
Further we know that triangles of the same altitude are
to one another as their bases, and also that triangles of equal
bases are to one another as their altitudes. Hence we might,
as in the case of the rectangle, prove that the area of a triangle
varies as the product of the base and altitude when both base
and altitude vary.
VARIATION 293
Again, the volumes of cylinders of the same height are to
one another as their bases, and also the volumes of cylinders
with equal bases are to one another as their heights. Hence
it might be proved that the volume of a cylinder varies as the
product of the base and height, when both base and height
vary.
From these illustrations we infer the general theorem: —
If A varies as B when C is constant, and A varies as C when
B is constant, then A varies as BC when both B and C vary.
Definition — One quantity is said to vary jointly as a number
of others when it varies directly as their product.
Example 1:
The volume of a cone varies jointly as its altitude and the
area of its base. The volume is 3927 cu. in. when the alti
tude is 15" and the diameter of the base 10". Find th.e
diameter when the altitude is 22" and the volume 436
cu. in.
Here Foe AB or V = kAB.
Substituting the first conditions:
3927 = * 15X7854X10 2 .
giving fe 15x . 7864xl0 r
From the second conditions:
436= 1 5 X 3 7854 7 XW X22B 
436X15X7854X10 2
22X3927
If d be required diameter,
436X15X7854X10 2
then 7854 d 2 =
.'. d* =
22X3927
436X15X10 2
22X3927
d = &7".
294 MATHEMATICS FOR TECHNICAL SCHOOLS
Example 2:
The volume of a gas varies inversely as the pressure and
directly as the absolute temperature (the absolute tempera
ture is obtained by adding 273 to the temperature on the
Centigrade scale).
If a quantity of nitrogen under 900mm. pressure at 20° C.
occupies a volume of 300cc, what volume will it occupy at
100° C. under 600mm. pressure?
T T
Here, V ex p or V = k p.
From first conditions:
onn . 293 . . . 300X900
300 = fc 9 ^, giving Jc = 293
From second conditions:
T , 300X900 373 __« _
F = 293" X 600 = 572  87cC 
Exercises CXIII.
1. The area of a triangle varies jointly as its base and alti
tude. The area of a triangle whose base is 19' and whose
altitude is 10' is 95 sq. ft. Find the altitude when the base
is 225' and the area 134 sq. ft.
2. The volume of a pyramid varies jointly as its height
and the area of its base. When the height is 18' and th(> base
a square 8' to the side, the volume is 384 cu. ft. What is
the side of the base if a pyramid of the same form, 10' high,
has a volume of 432 cu. ft.?
3. The pressure of the wind perpendicular to a plane surface
varies jointly as the area of the surface and the square of the
velocity of the wind. Under a velocity of 16 miles per hour
the pressure on 1 sq. ft. is 1 lb., what is the velocity when
the pressure on 3 sq. yd. is 68 lb.?
4. The amount of illumination received by a body varies
directly as the intensity of the light and inversely as the square
of the distance from the light.
VARIATION 295
From a light of 14 candlepower, the illumination is 5 at
a distance of 8'. Find the illumination at a distance of 10'
from a light of 40 candlepower.
5. The intensity of a magnetic field varies directly as the
number of vibrations and inversely as the square of the dis
tance of the magnet. When the distance is 5", and the
number of vibrations 15, the intensity is • 14. Find the
intensity when the distance is 075", and the number of
vibrations 90.
6. The heat developed in a conductor varies jointly as the
resistance of the conductor, the time the current flows, and
the square of the current. In 2 minutes a current of 4 amperes
developed 1400 units of heat in a wire having a resistance of
11 ohms. Find the resistance of a wire of the same size in
which 20,000 units of heat were developed by a current of
684 amperes in 3£ minutes.
7. The stiffness of a rectangular beam varies jointly as the
breadth and as the cube of the depth. Show that two beams
of the same material of breadth and depth (1) 2", 3", (2) 1"
378", are of nearly the same stiffness.
8. If the attraction between two masses varies directly as
their product and inversely as the square of the distance
between their centres, what would 1 lb. weigh on the surface
of a planet of the same density as the earth, but 1 • 5 times the
diameter?
9. The square of the time which a body takes to slide down
an incline varies as the square of the length and inversely as
the height. If the time taken is 1 sec. when the height is
4' and length 8', what must be the height of a plane 3' long
so that the body may slide down in \ sec?
10. The volume of a cylinder varies jointly as its
base and height. Of two cylinders volume of 1st: volume
of 2nd = 11: 8, and height of 1st: height of 2nd = 3:4.
If the base of the first is 165 sq. ft., find the base of the
second.
11. The volume of a gas varies inversely as the pressure
and directly as the absolute temperature.
What would be the volume at 0°C. and 760mm. pressure
of a mass of oxygen whose volume is 60cc. at 40C, under
a pressure of 750mm. of mercury?
296 MATHEMATICS FOR TECHNICAL SCHOOLS
12. At what temperature must a gas be so that its volume
will be 15 litres when the pressure is 800mm., if its volume
is 175 litres when its temperature is 100°C, and the pressure
700mm.?
13. The electrical resistance of a wire varies directly as
the length and inversely as the square of the diameter of the
wire. Its weight varies jointly as the length and the square
of the diameter.
If a pound of wire of diameter 06" has a resistance of
•25 ohms, what is the resistance of a pound of wire of the
same material, the diameter being 01"?
CHAPTER XXL
GEOMETRICAL PROGRESSION.
184. Amount. If I deposit $100 in a savings bank which
pays interest annually at 4%, I will be entitled to $4 interest
at the end of the first year. If I choose to leave this interest
on deposit my bank account would then be $104. This sum,
representing the principal plus the interest, is said to be the
amount of $100 in one year.
Consider the following examples.
Example 1:
Find the amount of $100 in 3 years at 6% per annum, com
pounded yearly.
The interest at the end of the first year = — of $100.
The sum itself = j52 f $100.
.'. the amount at the end of the first year= ^ of $100.
= $100(106).
The interest at the end of the second year = ^r of $100(1 • 06).
.'. the amount at the end of the second year
= j^ of $100(106).
= $100(1 06) 2 .
The interest at the end of the third year = — — of $100(1 06) 2 .
.. the amount at the end of the third year = pi $100(1 • 06) 2
= $100(1 06) 3
= $11910.
297
298 MATHEMATICS FOR TECHNICAL SCHOOLS
Example 2:
A man saves $200 a year for 4 years. If each year, he invests
it at 6% per annum, what are his accumulated savings 4 years
from the date of his first investment?
The amount of the first $200 = $200(1 06) 4 .
The amount of the second $200 = $200(1 06) 3 .
The amount of the third $200 = $200(1 06) 2 .
The amount of the fourth $200 = 8200(106).
Accumulated savings = $200(1 • 06) 4 + $200(1 06) 3
+$200(1 06) 2 +$200(l 06),
= $200(1 06 + 1 12360 + 1 19102 + 1 26248).
= $200X463710.
= $92742.
If in the preceding example the time had been 15 years,
instead of 4 years, it is apparent that considerable work would
be involved. The following discussion will develop a formula
for shortening the work.
Consider the Series:
a\ar\ar 2 \ar 3 ar n ~ l .
It is apparent that (1) each term is obtained from the pre
ceding by multiplying by r, called the common ratio, (2) in
the second term r is raised to the first power, in the third term
to the second power, .*. in the nth term to the (n — l) th power.
(3) there are n terms in the series.
Such a series is called a Geometrical Series and the terms in
the series are said to be in Geometrical Progression.
Let S = a + ar+ar 2 ar n  2 + ar n ~ l .
.'. rS= ar\ar 2 ar n ~ 1 \ar n .
.. rSS = ar n a.
.. S (rl)=a (r n l).
" S= r1
We here observe that in the above formula a is the first term,
r the common ratio, and n the number of terms.
GEOMETRICAL PROGRESSION 299
Returning to our previous difficulty in finding the accumu
lated savings at the end of 15 years, we have as in Example 2.
Accumulated Savings
= $200(1 06) 15 + $200(1 06) u . . . .$200(106)
= $200(1 06 + (l06) 2 ....(l06) 15 )}.
The series within the brackets is evidently a geometrical
series, the first term being 106, the common ratio 106,
and the number of terms 15.
. « , n J(l06) 15 l\ n J 239656l \.
..Sum = 1.06( 106 _ r ) = l06( 1Q6 _ 1 ^
. . . , a $200X1 06 XI 39656
.". Accumulated Savings = ^
• Ob
= $4934.88.
Note. — The value of (106) 15 may be obtained from interest
tables or by the use of logarithms.
185. Present Worth. If a person owes me $106 a year
from now, and money is worth 6%, I might as well accept
$100 now.
The $100 is here called the Present Worth of $106 and we
may, therefore, define the present worth of a future payment
as the sum which will at the given rate, amount to the pay
ment when due.
Example:
A man wills his son $1000 a year for 10 years. What is
this legacy worth now, if money is worth 5% per annum?
The amount of $1000 one year hence = $1000(1 05).
.•.81000(105) due one year hence has for present worth
SI 000.
$1000
.*. $1000 due one year hence has for present worth ' .
1 05
The amount of $1000 two years hence = $1000(1 05) 2 .
300 MATHEMATICS FOR TECHNICAL SCHOOLS
.". 81000(1 • 05) 2 due two years hence has for present worth
$1000.
$ 1 000
.'. $1000 due two years hence has for present worth^ — V,.
(1 05) 2
Similarly $1000 due three years hence has for present worth
$1000
(105) 3 '
.*. Present Worth of all the payments equals
$1000 $1000 , $1000 , $1000
105 _r (l05) 2_r (l05) 3 ^(l05) 10 '
$1000 fl + (l05) + (l05) 2 +(l05) 9,
$1000
"(105) 10
$1000^ 1628891 $1000 62889
\ io X i . ns' _ i ""n.n^MoX .n* $772154.
Pfi^}]
(105) 10/N 1051 ~(105) 10/ ^ 05
Exercises CXIV.
1. Find the amount of $450 if left on deposit in a bank for
3 years, if interest at 3% per annum compounded halfyearly
be allowed?
2. What sum of money loaned at 5% per annum will in 7
years yield $40710 interest?
3. What sum will amount to $198686 in 17 years at 4%
per annum?
4. A person deposits $100 in a savings bank on January
1st, 1911, and the same sum each year until January 1st, 1921.
If banks pay 3% per annum, compounded halfyearly, what
sum stands to his credit just after making the deposit on
January 1st, 1921?
5. A person holds $6000 in bonds paying 5% per annum.
He dies leaving the income for the first 10 years to his son.
If money is worth 6% per annum, what is the present worth
of the legacy?
6. A mortgage for $5000 with interest at 6% per annum
has 5 years to run. It is necessary to realize on the mortgage.
What sum should a person pay for it, if he wishes to make 7%
on his money?
Under what conditions would it be worth $5000?
MISCELLANEOUS EXERCISES 301
7. On November 1st, 1920, Brown invests $6000 in Victory
Bonds, due November 1st, 1933, paying 5f% per annum
payable halfyearly. If money is worth 6% per annum
compounded halfyearly, what is the present worth of the
bonds? (November 1st, 1921).
8. A man deposits $200 a year with a loan company which
pays 4% per annum compounded quarterly. What sum
stands to his credit at the end of 5 years?
9. A man dies leaving an annuity of $500 to his eldest son
for 10" years and then to his second son for the following 10
years. If money is worth 6% per annum, what is the present
worth of each legacy?
10. A municipality borrows $60,000 at 5% per annum.
What amount must be collected each year so that the debt
may be discharged in 10 equal annual payments, if money
is worth 6% per annum?
11. A man invests $500 in a business which pays 5% per
annum. Each year he invests an amount 10% greater than
the previous year. What amount stands to his credit at
the end of 10 years, if he reinvests his dividends in the
business?
12. A man takes a 20year endowment policy of $1000 on
which the annual premium is $4850. If he dies just after
the twelfth payment, how much more will his heirs receive
than if he had invested the money at 5% per annum? If he had
lived, how much less will he receive than if he had invested
the money as above?
13. A man takes a straight life policy for $5000 on which
the annual premium is $136. If he dies just before making
the 25th payment, compare the financial returns with having
invested the premium each year at 6%.
MISCELLANEOUS EXERCISES.
1. A man ordered 11^ tons of coal. The first load contained
9500 lb., the second 7000 lb. How many tons remain to
be delivered?
2. A tank containing 400 gallons has two pipes opening
from it. One pipe can empty it in 2 hours, the other in 2\
hours. If both^ pipes be opened for 15 minutes, how many
gallons are left in the tank?
302 MATHEMATICS FOR TECHNICAL SCHOOLS
3. The recent summer vacation extended for 72 days. A
boy spent ^ of it in the country, f of it camping, and the
remainder in the city. How many days did he spend in the
city?
4. A bricklayer received 90c. an hour for an 8hour day.
In the last year he worked 221 days, what was his total income?
5. A gang of men working on the roadway place 24 cu.
yd. of concrete in 1 hour. How many cu. yd. do they place
in 3 days of 8^ hours each?
6. An alloy contains ^f copper, \ tin, and the balance zinc.
How many lb. of each are there in an alloy of 336 lb.?
7. If beef is worth 21f c. per lb., what is the value of beef
weighing 532 lb.?
8. One pipe can empty a tank in 3 hr., and another pipe
can empty it in 2f hr. In what time can they both empty
it if running together?
9. A drill with a feed of 01" per revolution is making
80 R.P.M. How long will it take to drill 25 holes through a
\" plate if 15 seconds be lost in setting for each hole?
10. A contractor is to finish a piece of roadway in 20 days.
Twelve men work for 8 days and do \ of the work. How many
men must be employed for the balance of the time to finish
according to contract?
11. A certain pump delivers 143 gallons per stroke. If a
gallon of water weighs 10 lb., what weight of water will be
delivered in 218 strokes?
12. A drill with a feed of 100 is making 50 R.P.M. If \
of the time is used in setting, how many holes can be drilled
in a \" plate in 2 hours?
13. A gallon of water contains 277274 cu. in. What is
the percentage error in taking 6j gallons as equivalent to 1
cu. ft.?
14. A reamer 9" long is 1375" in diameter at one end and
1*125" at the other end. What is the taper per foot?
15. How long will it take to excavate to a depth of 4' for
a building having a frontage of 74' 0", and a depth of 243' 0",
using a steam shovel if the bucket holds twothirds of a cu.
yd. and is filled 3 times every 2 minutes?
16. The American gallon contains 231 cu. in. What per
cent, is it of the English gallon?
MISCELLANEOUS EXERCISES 303
17. In building a roadway containing 1000 cu. yd., the
total cost was as follows: — 4000 bags of cement at SI. 20;
1000 cu. yd. of stone at $2.00; 450 cu. yd. of sand at $1.80;
lumber $225; labor $3240, tools, etc., $200. Find the average
cost per cu. yd.
18. A pressure of 485 lb. per sq. ft. is how many ounces
per sq. in.?
19. A man walks 2\ miles. If his average step is 2' 7",
how many steps does he take?
20. A room is to be floored with £" tongued and grooved
flooring taken out of 6" material, allowing §* in width for
the dressing and tongue. The room is 12' 6"X15' 6", and
has a square bay window 10' 3" wide X 3' 0" deep. How
many square feet are in the room and what number ot feet
run of flooring would be required?
21. Reduce 13 quarts to the decimal of a bushel.
22. If sound travels 1120 ft. per sec, how long will it take
to hear the report of a gun fired at a distance of 8 miles?
23. An electriclight bill was 16f% higher this month than
last. It was $3.42 last month, what is it this month?
24. In a certain machine f of the power supplied is lost in
friction, etc. What is the percentage efficiency of the machine?
25. In a room 15' 3"X22' 6" it is required to lay a clear
quarter cut white oak floor f'Xl^" face measure. Allowing
30% for loss in dressing and working tongue, how much would
the flooring cost at $275 per thousand sq. ft.?
26. In a vernier caliper the reading shows 1", 2 tenths, 1
small division, while the 8th division of the vernier is in line
with a beam division. In reading the instrument I neglect to
add the vernier reading. What is the percentage error?
27. A steel hook when immersed in water displaced 1
quarts. If the specific gravity of steel is 78, find the weight
of the hook.
28. When casting iron pipe, an allowance of f " per foot is
allowed for shrinkage. What percent, is this?
29. If the specific gravity of ice is 921 and of salt water
1024, what part of an iceberg is below the surface of the
water when floating?
30. An alloy contains 225% of nickel and 12% of carbon.
How many pounds of each are there in 500 pounds of the
alloy?
304 MATHEMATICS FOR TECHNICAL SCHOOLS
31. Find the cost, at $60 per M, of building a walk 60'
long by 6' wide; plank to be 2" thick and laid crosswise on
3 pieces, 4"X4".
32. A mixture for casting contains 5 parts copper, 4 parts
lead, and 3 parts tin. What is the percentage composition?
33. Find the cost of shingling a roof 16'X20' with shingles
laid 4" to the weather, if the cost of material and labour is
$12 per square of shingles.
34. A road bed rises 2' 3" in 300'. What percent, grade
is this?
35. Find the length of the . longest straight line that can
be drawn in a room 20' long 16' wide and 12' high.
36. A barn is 40' 0" wide and 60' 0" long, with a roof £
pitch. The rafter heel and projections at ends are each 2'.
Find the cost of covering with 1 " sheeting at $60 per M,
allowing 5% for waste.
37. Find the weight of a hexagonal bar of iron " to a side
and 6' long. (1 cu. in. = 26 lb.).
38. Find the size of tap drill for a &*, 12 pitch, sharp "V"
thread nut.
39. An equilateral triangle has an area of 3254 sq. in.
What is the length of a side?
40. In a steel plate 4'X2' 3" and \" thick, 10 round holes
are bored each \\" in diameter. Find the weight of the plate
after boring.
41. The diameter of the safety valve in a boiler is 3^".
If the pressure of the steam is 150 lb. per sq. in., find the
total pressure on the valve.
42. A well is to be sunk 4' in outside diameter and 24' deep.
If the carts used for carting away the earth hold 1^ cu. yd.,
and excavated earth increases in bulk 15%, how ninny cart
loads will there be? A concrete slab 8" thick is placed on
the bottom, and the sides are bricked, using 4 bricks per super
ficial foot. How many cu. yd. of concrete would be used
and how many bricks?
43. The driving wheels of a locomotive are 4' in diameter,
and the speed of the locomotive is 40 miles per hour. How
many revolutions must the drivers make per minute?
44. A tank 7' in diameter is bound by 4 wroughtiron hoops
2" wide and i§" thick. Find their weight.
MISCELLANEOUS EXERCISES  305
45. The maximum speed for an emery wheel is a mile a
minute. Find the maximum number of revolutions per sec.
for an emery wheel 8" in diameter.
46." A semicircular platform with a diameter of 18' has a
table 2' wide around its semicircumference. Find the area
of the table and the available space for seating accommodation.
47. The length of the shadow of a 2' rod is 1' 6" and at the
same time the shadow of a tree is found to be 30'. Find the
height of the tree.
48. What is the offset of the tail stock for turning a taper
18" long on a bar 30" long if the diameters at the ends of the
taper are 3$ " and 2\ " ?
49. Find the size of the largest square timber which can be
cut from a log 18" in diameter.
50. Find the diameter of a tap drill for a Whitworth nut
for a screw of outside diameter If", double threaded and 14
pitch.
51. Three circles each of radius 6" are enclosed in an equi
lateral triangle; find the side of the ^triangle.
52. In riding a certain bicycle one revolution of the pedals
gives two revolutions of the wheels. If the wheels are 26"
in diameter, how many revolutions of the pedals per min.
will give a speed of 12 miles per hour ?
53. The diameter of a cylindrical winch barrel for a crane
is 10". If a rope with a diameter of f " be used, how many
coils of rope and what length of barrel would be necessary to
raise a load 30' ?
54. If the speed of a point on the circumference of a fly
wheel must not exceed 5000 ft. per min., find the maximum
diameter for the wheel in order to make 120 R.P.M.
55. A steel bar of square crosssection is to be equal in area
to a rod 6" in diameter. Find a side of the square.
56. A triangular steel plate is to have its sides 13", 14",
and 15", and to weigh 298 lb.; find its thickness.
57. An elliptical steel plate has a major axis of 12" and a
minor axis of 10". It is f " thick; find its weight.
58. A cylinder 14" in diameter fits in a cubical box. Cal
culate the percentage void.
59. The commutator of a dynamo is 24" in diameter and
16" long. Find the radiating surface in sq. ft. (lateral sur
face).
306 MATHEMATICS FOR TECHNICAL SCHOOLS
60. Find the diameter of a circular plate equal in area
to an elliptical plate major and minor axis 18" and 12"
respectively.
61. The shaft of a squareheaded bolt is 1" in diameter,
the head being f " thick and l\" to the side. Determine the
length of the shaft to have twice the weight of the head.
(1 cu. in. = 26 lb.).
62. Find to the nearest sixteenth of an inch the length of
a " steel rod that is turned per min., if the cutting speed is
40 ft. per min. and the feed 24.
63. The length of the core of a dynamo is 20" and it must
have a radiating surface of 500 sq. in. Find its minimum
diameter.
64. An elliptical funnel has a major axis of 18' and a minor
axis of 12'. Find the discharge of smoke in cu. ft. per min.,
if at a rate of 12 ft. per sec.
65. The pulley on the armature shaft of a dynamo is 3\"
in diameter. This is belted to a driving shaft which makes
400 R.P.M. The speed of the dynamo must be 1800 R.P.M.
What sized pulley must be placed on the driving shaft ?
66. Find the weight of a coil of copper wire 400' long,
if the area of the crosssection of the wire is 40,000 circular
mils. (1 circular mil = area of a circle one mil, or 001", in
diameter).
67. Two hundred and fortyfive sq. ft. of zinc are required
in lining the sides and bottom of a cubical vessel. How
many cu. ft. of water will it hold?
68. Four concrete abutments are to be built for a bridge.
Each abutment is to be 8' 0" high, 3' 0" thick, 18' 0" long at
the bottom and 12' 0" at the top. If one cu. yd. of concrete
requires 25 cu. ft. of stone, 12 cu. ft. of sand, and 4 bags of
cement (1 bag=l cu. ft.), find the quantity of each necessary
for the job.
69. Find the total cost of building a rubble stone wall for
a basement 30' 0" by 26' 0" by 8' 0" high, the wall being 18"
thick, if the stone costs $60 a toise and the labour including
sand and lime, is 30c. per cu. ft. (Exact length of wall is
taken with no allowance for openings).
70. It is required to build a brick wall on the front and one
side of a lot 35' 0" in frontage and 128' 0" in depth. The
wall is to be 7' 0" high and single brick 9" in thickness There
are two gates one 12' 0"X7' 0" and the other 3' 0"X7' 0".
MISCELLANEOUS EXERCISES 307
Piers for strengthening the wall add 12% to its cubic contents.
Allowing 15 bricks per cu. ft., at $18 per thousand, and $12
per thousand for laying, find the total cost.
71. A vessel to hold 10 quarts is to have an elliptical cross
section with major and minor axis 12" and 8" respectively.
Find the height and the number of sq. ft. of tin in the vessel.
72. A garage 10' 0" wide and 16' 0" long is to have a gable
roof,  pitch. The following material is supplied:
Rafters 2"X4", 2' on centre, 18" heel— $65 per M.
Sheeting, 1" thick, projecting 6" on ends — $63 per M.
Shingles laid A\ " to the weather at $9 • 50 per square.
Find the cost of the above material.
73. In drilling in mild steel a 1£" twist drill makes 40
R.P.M., with a feed of 72. How many cu. in. will be cut
away in 5 minutes?
74. A chimney is to be built in a residence with the following
dimensions: height from basement floor to first floor level is
9' 0"; from first to second floor level 10' 6"; from second to
third floor 9' 6"; and from third floor to top of stack 15' 0".
The size of the stack from basement to second floor level is
T 6" by 2' 7"; from second floor to third floor 4' 10" by 2' 7";
from third floor up 4' 1" by 2' 7". From basement to top of
stack a 1' 1" by 1' 1" furnace flue is run. On first floor a
fireplace opening 3' 0" by 2' 6" by 1' 10" deep is to be built.
From top of fireplace opening a 9" by 1' 1" flue is carried up
to top of stack. On the second floor a fireplace opening
1' 10" by 2' 6" by 1' 1" deep is to be built having a 9" by 9"
flue carried from top of opening to top of stack. Allowing
15 bricks to the cu. ft., how many bricks would be necessary
to build the stack ?
75. Find the number of minutes required for turning a
shaft 5" in diameter and 6' long, the cutting speed being
40 ft. per min. and the feed 100.
76. Two wheels 8" and 6" in diameter are running on parallel
shafts 4' apart. Find the length of an open belt connecting
the two wheels. (1) Using the formula S±(R + r)+2d. (2)
Using the' exact method. Hence find the percentage error
in the formula.
77. The circumference of the base of a church spire in the
form of a cone is 42' and the height is 80'. Find the cost of
covering with sheetiron at 30c. a sq. ft.
308 MATHEMATICS FOR TECHNICAL SCHOOLS
78. A chimney shaft 70' 0" high is to be erected having a
flue averaging 3' 0"X3' 0" from bottom to top. The shaft
is square and for the first 14' 0" the walls are 1' 10" thick;
the next 14' 0" is 1' 6" thick; the next 20' 0" is 1' 1" thick
and the remaining portion 9" thick. How many bricks
would be required allowing 15 bricks to the cu. ft.?
79. How many sq. ft. of sheetiron will it take to roof a
hemispherical dome 30' in diameter?
80. A building 24' 0" wide and 36' 0" long is to have a hip
roof, I pitch, with an 18" overhang, measured horizontally
(formed by extending the rafters). Find the cost of the
following material:
Hip Rafters, 2 // X6' / $55 per M.
Rafters, 2"X6"(2' on centre), §55 per M.
Square sheeting, §" thick (10% added for cutting),— $53
per M sq. ft.
Slate, gauge 7\" , — $30 per square.
81. A hollow copper sphere used as a float weighs 1 lb.,
and is 6" in diameter. How heavy a weight will it support
in the water?
82. Grain dumped in a pile makes an angle of 30° with the
horizontal. How many bushels will there be if the pile forms
a regular cone 10' in diameter?
83. A tank 10' long and 2' in diameter is in the form of a
cylinder with hemispherical ends. How many gallons will
it hold?
84. A steel pin 6" long and 1" in diameter at the large end
has a B. & S. taper. Find its weight.
85. A chimney shaft 80' 0" high is to be erected, having a
flue averaging 3' 0" in diameter from. bottom to top. The
shaft is circular and for the first 15' 0" the wall is 2' 0" thick,
the next 15' 0" is 1' 6" thick, the next 25' 0" is 1' 0" thick, and
the remainder is 9" thick. How many bricks will be required
allowing 15 bricks per cu. ft.?
86. An oil tank in the form of a cylinder 15' long and 3'
in diameter is lying on its side. It is filled to a depth of 30".
How many gallons of oil does it contain and what is the sur
face of the tank not in contact with the oil?
87. A ring of outer diameter 16" is made of round cast
iron is" in diameter. Find its total surface and weight.
MISCELLANEOUS EXERCISES 309
88. A water pail has a base 12" in diameter and a top 16"
in diameter. The height of the pail is 18". Find the capacity
>n gallons and the sq. ft. of material used in construction.
89. A spkere 8" in diameter is penetrated axially by a
cylindrical hole 4" in diameter. Find the volume of the
remaining solid.
90. A tank is in the form of a cylinder with segments of
spheres for ends. The total length is 8', the cylindrical part
7', and the diameter 2'. Find the capacity in gallons.
91. A building 24' 0" wide and 40' 0" long is to have a hip
roof, ^ pitch, with a 2' overhang, measured horizontally (formed
by extending the rafters). Find the cost of the following
material :
Hip rafters, 2" X6" $60 per M.
Rafters, 2" X6"(2' on centre), $50 per M.
Square sheeting, 1" thick (8% for cutting), — $45 per M.
Shingles, 4j" to the weather, at $9.50 per square of shingles.
92. In a room 16' 4" by 20' 8" it is required to lay a quarter
cut clear white oak floor 3"X1§", face measure. Allowing
30* { for loss in dressing and working tongue, find the cost at
$250 per M square feet.
93. A lifebuoy elliptical in crosssection has major and
minor diameters of 5" and 3" respectively. If the mean
diameter be 30", find the volume in cu. in.
94. A conical tent 9*' high is to be of such a size that a man
6' high can stand erect anywhere within 3' of the centre pole.
How many yd. of canvas 27" wide does it contain?
95. Find the weight of a sheet of metal weighing 650 oz.
per sq. ft., if the equidistant half ordinates at 15' intervals
are 0, 175, 225, 3, 425, 635, 6 ft.
96. Two wheels 9" and 8" in diameter are running on parallel
shafts 5' apart. Find the length of a crossed belt connecting
the two wheels. (1) Using the formula 3§(# + r)+2d. (2)
Using the exact method. Hence, find the percentage error
in the formula.
97. A room 12' 0" wide and 17' 0" long is to be floored with
No. 1 quality birch, " X21", to cost 30c. a sq. ft. ; 33% being
added for dressing and working the tongue. It is also to be
paneled to a height of 4' 0" at a cost of 80c. a sq. ft. There are
2 doors each 2' 8" wide and 4 windows each 2' 6" wide, the sills
being 2' 0" from the floor. Find the total cost.
310
MATHEMATICS FOR TECHNICAL SCHOOLS
98. A lot 50' 0" X160' 0" is to be enclosed by a picket fence.
The pickets are 4' 0" long, 3" wide and 1" thick, and are placed
3" apart. The posts are placed 8' 0" apart, scantling 2" X
4" being used at top and bottom for railing, and a baseboard
10" wide. If the posts cost 30c each, the lumber $52 per M,
and the pickets $10 per hundred, find the total cost of material.
99. With a feed of \" per revolution, how fast is it neces
sary to run a bar, to turn 40" long in 10 minutes?
Fig. 131
100. The above figure represents a crosssection of an egg
shaped sewer. OE is the right bisector of AB and equal in
length to AB. The semicircular top has a radius 0.4 = 12".
The sides are arcs with radii CB and DA each equal 36".
The small end is an arc with radius FE = Q>". Find the area
of the crosssection.
MISCELLANEOUS EXERCISES.
1. The area of a circle is irr 2 ; express the diameter in terms
of the area.
2. Express the area of a rectangle in terms of its perimeter
when the length is twice the width.
MISCELLANEOUS EXERCISES 311
3. Three circles are to touch one another and have their
centres 3", 4" and 5" apart. Find the diameters of the
circles.
W ird 2
4. The stress in a tie bar is /=r where A = j~. Find
d when/ = 7000 and JF = 2100.
5. The difference of the acute angles of a rightangled
triangle is 15°. Find the angles.
6. The weight of a body varies inversely as the square of
its distance from the centre of the earth. If a man weighs
190 lb. at the earth's surface, what would he weigh on the
top of a mountain 4£ miles high (radius of earth = 4000 miles).
4
7. The volume of a sphere is ^t 3 . Find the radius when
o
the volume is 616 cu. in.
A
8. The stress in a beam is given by/ = ,— . Find A when/ =
cd 2
6000, «=p c = 3, d = 5.
9. The time of vibration of a pendulum varies as the square
root of its length, for a given latitude. A pendulum 39 • 1"
long vibrates once per second. What is the length of a pendu
lum that vibrates three times per second?
10. An exterior angle at the base of an isosceles triangle
is 108°; find all the angles.
11. The velocity of flow of water under a head h is V —
ay/2gh. Find h when V = 36, g = 32, a = • 5.
12. In an isosceles rightangled triangle the perp. from the
vertex on the hypotenuse is 8", find all the sides.
13. The twisting moment on a solid circular shaft is given
by T = ^. Find d when T= 150000, / = 6000.
14. A chord 8" long is 6" from the centre of a circle. Find
the radius of the circle.
15. The moment of inertia of a body about an axis is given
7 2
by I = — . Find k when I = • 5, w = 38, a = 32.
9
16. The radius of a circle is 16". How far from the centre
is a chord 8" in length?
312 MATHEMATICS FOR TECHNICAL SCHOOLS
17. The law of a machine is given by P = x+yw. Find x
when P = 648, y= 2, w = 62.
18. The length and breadth of a rectangular floor differ
by 6'; the area is 72 sq. ft., find the perimeter.
19. An equilateral triangle has sides 20" in length. Find
the radius of the inscribed and circumscribed circles.
20. The coefficient of selfinduction of a coil of wire is
. T 4ttAti 2
given by L= /1Q> .
Find n, when A = irr 2 , r = 25, Z=015, Z = 40.
21. The difference between the lengths of the parallel sides
of a trapezium is 4; the area is 100, and the sum of the parallel
sides 20. Find the dimensions.
22. The lifting power of an electromagnet is given by
8x '
P being the pull in dynes. Find P, when B = 14000, A = 30.
23. I have to walk a distance of 144 miles, and I find that
if I increase my speed by 1^ miles per hour I can walk the dis
tance in 14 hours less than if I walk my usual rate. Find my
usual rate.
24. In measuring a rectangle the length is measured \\%
too small and the width 2% too large. Find the percentage
error in the area.
25. The perpendicular from the vertex of the right angle
of a rightangled triangle to the hypotenuse is 2§". The
hypotenuse is 5". Find the other sides.
26. The length of a cylinder is twice its diameter. Find
the diameter so that the cylinder may contain three times as
many cu. ft. as a sphere 6" in diameter.
27. A ton of lead is rolled into a sheet \ " thick. Find the
area of the sheet if a cu. ft. of lead weighs 712 lb.
28. A rectangular piece of tin has an area of 195 sq. in.,
and its perimeter is 56". Find its dimensions.
29. Around the outside of a square garden a path 3' wide
is made. If the path contains 516 sq. ft., find a side of the
garden.
MISCELLANEOUS EXERCISES 313
30. An open box is made from a square piece of tin by
cutting out a 2" square from each corner and turning up the
sides. How large is the original square if the box contains
1152 cu. in.?
31. A man whose eye is 5' 6" above the ground, sights over
the top of a 12' pole and just sees the top of a tower. If he is
7' from the pole and 63' from the tower, find the height of
the tower.
32. A derrick for hoisting coal has its arm 24' long. It
swings over an opening 20' from the base of the arm. How
far is the top of the arm above the opening?
33. Show that the area of a triangle is y/s (s — a) (s — b) (s — c) ,
where a, b, and c are the sides and s half their sum.
34. An open box is made from a rectangular piece of tin
twice as long as it is wide, by cutting out a 2" square from each
corner and turning up the sides. If the total surface is 56
sq. in., find the dimensions of the original piece.
35. At a school entertainment the price of the tickets
for the second performance was reduced 20%, which resulted
in an increase in receipts of 10%. What was the percentage
increase in the number of tickets sold ?
36. Two branches of an iron water pipe are respectively
\\" and 2\" in diameter. Find the diameter of a pipe that
will just carry away the water from both branches.
37. If a man spent \ of his salary for board, \ of the remainder
for other expenses, and saved annually $400, what was his
salary?
38. The resistance offered by the air to the passage of a
bullet through it varies jointly as the square of its diameter
and the square of its velocity. If the resistance to a bullet
whose diameter is 25" and whose velocity is 1600' per second
is 485 oz., what will be the resistance to a bullet whose
diameter is 4" and whose velocity is 1550' per second?
39. The sides of the base of a triangular prism are as 3:4:5,
and its volume is 270 cu. in. If the altitude is 5", find the
sides of the base.
40. The ends of a frustum of a cone are respectively 8"
and 2" in diameter. If the lateral surface is equal to the area
of a circle whose radius is 5", find the height of the frustum.
314 MATHEMATICS FOR TECHNICAL SCHOOLS
41. The safe distributed load in wood beams is given by
c\by.d 2
S.L. = j , where c is a const., b the breadth in in.,
d the depth in in., L the length in ft.
Solve for c, b, d and L.
If c = 100, 6 = 10", d = 6", Z = 10' find S.L.
42. The horsepower required to drive air through a pipe
Q 3 L
is given by H.P. = J~ „ „ , where Q is the volume in cu. ft. per
sec., L the length in ft., d the diameter in in.
Solve for Q, L and d.
If Q = 10 cu. ft., L = W, d = 10", find H.P.
TABLES
315
Decimal Equivalents of Parts of an Inch.
67'
1
32 .... .
3
"ST
1
TTT
5
■gT'
3
3~2' • ' •
7
"BT
1
¥
9
"BT
5
3~2'
11
3
TB
13
64
7
T2 • • •
15
"BT
T
17
"BT
19
64
A
•01563
•03125
•04688
•0625
•07813
•09375
• 10938
•125
• 14063
• 15625
•17188
•1875
•20313
•21875
•23438
•25
•26563
•28125
•29688
•3125
21
64
A3.
3 2
A5.
32
TS'
19
32
21
32'
23
64
25
6 1
2J7
64
29
^4
31
64
33
64
35
64
37
64
39
64
4JL
64
43
64
•32813
•34375
•35938
•375
•39063
•40625
•42188
•4375
•45313
•46875
•48438
•5
•51563
•53125
•54688
•5625
•57813
•59375
•60938
•625
•64063
•65625
•67188
•6875
45.
6 4 •
2 3
32
47
64 •
3
T
49
64 •
3 2
51
64 '
1 3
16
_51
6~4 •
2 7
32
55
64 •
T
8
5.1.
64 •
29
32
_5j>.
64 •
1 5
16
6 1 •
3 1
32
63
64 •
1 1
•70313
•71875
• 73438
•75
•76563
•78125
•79688
•8125
•82813
•84375
•85938
•875
•89063
•90625
•92188
•937.5
•95313
•96875
•98438
• 00000
316
MATHEMATICS FOR TECHNICAL SCHOOLS
The following Tables are from Kent's Engineers' Pocket
Book:
Weight and Specific Gravity of Stone, Brick, Cement, Etc.
Specific
Gravity
Weight in lbs.
per cubic foot
Asphaltum
139
16
179
20
216
224 to 24
218
16
179
87
Brick, Soft
" Common
100
112
" Hard
125
" Pressed
135
" Fire
140 to 150
" SandLime
136
Brickwork in Mortar
" in Cement .......
Portland Cement (loose) ....
100
112
92
(in barrel) .
Clay
115
192 to 24
192 to 248
1 • 15 to 1 • 28
144 to 176
256 to 272
16 to 192
•8 to 96
23 to 29
256 to 288
224 to 256
224 to 268
144 to 16
115
150 to 181
264
144 to 176
224 to 24
272 to 288
216 to 34
176 to 192
120 to 150
Concrete
120 to 155
Earth, loose
72 to 80
" rammed
90 to 110
Granite
160 to 170
Gravel
100 to 120
Lime, Quick, in bulk
Limestone
50 to 60
140 to 185
Marble
160 to 180
Masonry, dry rubble
M dressed
140 to 160
140 to 180
Mortar
90 to 100
Pitch
72
Plaster of Paris
93 to 113
Quartz
165
Sand
90 to 100
Sandstone
140 to 150
Slate
170 to 180
Stone, various
135 to 200
Tile
110 to 120
TABLES
317
Weight and Specific Gravity of Wood.
Specific
Gravity
Mean Value
Weight in lbs.
per cubic foot
Ash
Bamboo
Beech
1
72
35
73
65
62
66
56
53
23
61
59
38
81
68
77
74
45
61
48
45
82
58
54
45
22
46
Birch
41
Cedar
39
Cherry
41
Chestnut
Cypress
35
33
Ebony
76
Elm
38
Fir
37
Hemlock
Mahogany
Maple
24
51
42
Oak, White
Oak, Red
Pine, White
Pine, Yellow
Poplar
48
46
28
38
30
Spruce
28
Teak
51
Walnut
36
Willow
34
318 MATHEMATICS FOR TECHNICAL SCHOOLS
Weight and Specific Gravity of Metals.
Aluminum
Brass : — Cu. + Zn
'80 20^
70 30
60
,50
40
50J
Bronze
Copper
Iron, Cast ....
Iron, Wrought
Lead
Magnesium. . .
Mercury
Nickel
Platinum
Silver
Steel
Tin
Tungsten
Zinc
Specific
Gravity
Mean Value
267
86
840
836
82
8853
8853
7218
77
1138
175
136
88
215
10505
7854
735
173
700
Weight in lbs.
per cubic foot
1665
Weight in lbs.
per cubic inch
0963
5363
•3103
5238
•3031
5213
■3017
5114
•2959
552
•3195
552
•3195
450
•2604
480
•2779
7097
•4106
109
•0641
848
•4908
5487
•3175
1347
•7758
655 • 1
•3791
4896
•2834
4583
•2652
10787
•6243
4365
•2526
LOGARITHMS
Logarithms.
319
Mean Differences
1
2
3
4
5
6
7
8
9
1 2
3
4
5 6
7 8 9
10
ooooi i
00432
00860
01284
01703
42 85 127
170 212 254
297 339 381
02119
02531
02938
03342
03743
40 81
121
162 202 242
283 323 364
II
04139
04532
04922
05308
05690
37 77 116
154 193 232
270 309 348
06070
06446
06819
07188
07555
37 74 111
148 185 222
259 296 333
12
07918
08279
08636
08991
09342
36 71
106
142 177 213
248 284 319
09691
10037
10380
10721
11059
34 68 102
136 170 204
238 272 307
13
11394
11727
12057
12385
12710
33 66
98
131
164 197
229 262 295
13033
13354
13672
13988
14301
32 63
95
120
158 190
221 253 284
14
14613
14922
15229
15534
15836
30 61
91
122
152 183
213 244 274
16137
16435
16732
17026
17319
29 59
88
118 147 177
206 236 265
15
17609
17898
18184
18469
18752
28 57
86
114 142 171
199 228 256
19033
19312
19590
19866
20140
28 55
83
110
138 165
193 221 248
16
20412
20683
20951
21219
21484
27 53
80
107 134 160
187 214 240
21748
22011
22272
22531
22789
26 52
78
104
130 156
182 208 233
17
23045
23300
23553
23805
24055
26 50
70
101
126 151
176 201 227
24304
24551
24797
25042
25285
25 49
73
OS
122 147
171 196 220
18
25527
25768
26007
26245
26482
24 48
71
95
119 143
167 190 214
26717
26951
27184
27416
27646
23 46
00
03
116 139
102 185 208
19
27875
28103
28330
28556
28780
23 45
0s
00
113 135
158 180 203
29003
29226
29447
29667
29885
22 44
00
88
110 132
154 176 198
20
30103
30320
30535
30750
30963
31175
31387
31597
31806
32015
21 43
64
85
106 127
148 170 190
21
32222
32428
32634
32838
33041
33244
33445
33646
33846
34044
20 40
00
80
100 120
140 160 180
22
34242
34439
34635
34830
35025
35218
35411
35603
35793
35984
20 39
58
77
97 116
135 154 174
23
36173
36361
36549
36736
36922
37107
37291
37475
37658
37840
19 37
50
74
93 111
130 148 167
24
38021
38202
38382
38561
38739
38917
39094
39270
39445
39620
18 35
53
71
89 106
124 142 159
25
39794
39967
40140
40312
40483
40654
40824
40993
41162
41330
17 34
51
68
85 102
119 136 153
26
41497
41664
41830
41996
42.160
42325
42488
42651
42813
42975
16 33
40
66
82 98
115 131 148
27
43136
43297
43457
43616
43775
43933
44091
44248
44404
44560
16 32
47
63
79 95
111 126 142
28
44716
44871
45025
45179
45332
454S4
45637
45788
45939
46090
15 30
40
(il
76 91
107 122 137
29
46240
46389
46538
46687
46835
46982
47129
47270
47422
47567
15 29
44
50
74 88
103 118 132
30
47712
47857
48001
48144
48287
48430
48572
48714
48855
48996
14 29
43
57
72 86
100 114 129
31
49136
49276
49415
49554
40003
49831
49969
50106
50243
50379
14 28
41
55
69 83
97 110 124
32
50.51.5
50050
50786
50920
51054
51188
51322
51455
51587
51720
13 27
40
54
67 80
94 107 121
33
51851
51983
52114
52244
52375
52504
52634
52763
52892
53020
13 26
30
52
65 78
91 104 117
34
53148
53275
53403
53529
53656
53782
53908
54033
54158
54283
13 25
38
50
63 76
88 101 113
35
54407
54531
54654
54777
54900
55023
55145
55207
55388
55509
12 24
37
4!)
61 73
85 98 110
36
55630
55751
55S71
55991
56110
5022!)
5634s
56467
56585
56703
12 24
36
48
60 71
83 95 107
37
56S20
56937
57054
57171
57287
57403
57519
57634
57749
57864
12 23
35
40
58 70
81 93 104
38
57978
58092
58206
58320
58433
58546
58659
58771
58883
58995
11 23
34
45
57 68
79 90 102
39
59106
59218
59329
59439
59550
59660
59770
59879
59988
60097
11 22
33
44
55 66
77 88 99
40
60206
60314
60423
60531
60638
60745
60853
00050
61066
61172
11 22
33
44
55 66
77 88 99
41
61278
61384
61490
61595
61700
61805
61909
62014
62118
62221
10 21
31
42
53 63
74 84 95
42
62325
62428
62531
62634
62737
02 S3
62941
63043
63144
63246
10 20
31
41
51 61
71 82 92
43
03347
63448
63548
63649
63749
63846
63949
64048
64147
64246
10 20
30
40
50 60
70 80 90
44
64345
64444
64542
64640
64738
64836
64933
65031
65128
65225
10 20
20
39
49 59
68 78 88
45
65321
65418
65514
65510
65706
65801
65896
65992
66087
66181
10 19
20
88
48 57
67 76 86
46
66276
66370
66464
0055s
66652
66745
66839
66932
67025
67117
9 19
28
37
47 56
65 74 84
47
67210
07302
67394
67486
67578
67669
67761
67852
67943
68034
9 18
27
30
46 55
64 73 82
48
68124
6821S
68305
6S395
68485
68574
68664
68753
68842
68931
9 18
27
36
45 53
63 72 81
49
69020
69108
69197
6928S
011373
69461
6954s
69636
69723
69810
9 18
20
35
44 53
62 70 79
320
MATHEMATICS FOR TECHNICAL SCHOOLS
Logarithms.
Mean Differences
1
2
3
4
5
6
7
8
9
1 2 3
4 5 6
7 8 9
50
69897
BOOM
70070
70157
70243
70829
70418
70501
70688
70672
9 17 26
34 43 52
60 69 77
51
707.57
70842
70927
71012
71096
71181
71265
71349
71433
71517
8 17 25
34 42 50
59 67 76
52
7 1'IDi i
71684
71767
71850
71933
72016
72099
72181
72263
72346
8 17 25
33 42 50
58 66 75
53
72428
72509
72591
72673
72754
72835
72916
72997
73078
73159
8 16 24
32 41 49
57 65 73
54
73239
73320
73400
73480
73560
73640
73719
73799
73878
73957
8 16 24
32 40 48
56 64 72
55
74036
74115
74194
74273
74351
74429
74507
74586
74663
74741
8 16 23
31 39 47
55 63 70
56
74819
74896
74974
75051
75128
75205
75282
75358
75435
75511
8 15 23
31 39 46
54 62 69
57
76687
75664
75740
75891
75967
76042
76118
76193
76268
8 15 23
30 38 45
53 60 68
58
76343
76418
76492
76567
76641
76716
76790
76864
76938
77012
7 15 22
30 37 44
52 59 67
59
77085
77159
77232
77305
77379
77452
77525
77597
77670
77743
7 15 22
29 37 44
51 58 66
60
77815
77887
77960
78032
78104
78176
78247
78319
78390
78462
7 14 22
29 36 43
50 58 65
61
78533
78604
78675
78746
78817
78888
78958
79029
79099
79169
7 14 21
28 36 43
50 57 64
62
79239
79309
79379
79449
79518
79588
79657
79727
79796
79865
7 14 21
28 35 41
48 55 62
63
79934
80003
80072
80140
80209
80277
80346
80414
80482
S0550
7 14 20
27 34 41
48 54 61
64
80618
80686
80754
80821
80889
80956
81023
81090
81158
81224
7 13 20
27 34 40
47 54 60
65
81291
81358
81425
81491
81558
81624
81690
81757
81823
81889
7 13 20
26 33 40
46 53 59
66
81954
82020
82086
82151
82217
82282
82347
82413
82478
S2543
7 13 20
26 33 39
46 52 59
67
82607
82672
82737
82802
82866
82930
82995
83059
83123
83187
6 13 19
26 32 38
45 51 58
68
83251
83315
83378
83442
83506
83569
83632
83696
83759
83822
6 13 19
25 32 38
44 50 57
69
83885
83948
84011
84073
84136
84198
84261
84323
84386
84448
6 12 19
25 31 37
43 50 56
70
84510
84572
84634
84696
84757
84819
84880
84942
85003
85065
6 12 19
25 31 37
43 50 56
71
85126
85187
85248
85309
85370
85431
85491
85552
85612
85673
6 12 IS
24 31 37
43 49 55
72
85733
85794
85854
85914
85974
86034
86094
86153
86213
86273
6 12 18
24 30 36
42 48 54
73
86332
86392
86451
86510
86570
86629
86688
86747
86S06
86S64
6 12 18
24 30 35
41 47 53
74
86923
86982
87040
87099
87157
87216
87274
87332
87390
87448
6 12 17
23 29 35
41 46 52
75
87506
87564
87622
87679
87737
87795
87852
87910
87967
88024
6 12 17
23 29 35
41 46 52
76
88081
88138
88195
88252
88309
88366
88423
88480
88536
88593
6 11 17
23 29 34
40 46 51
77
88649
88705
88762
88818
88874
88030
88086
89042
89098
89154
6 11 17
22 28 34
39 45 50
78
89209
89265
89321
89376
89432
89487
89542
80507
89653
89708
6 11 17
22 28 33
39 44 50
79
89763
89818
89873
89927
89982
90037
90091
90146
90200
90255
6 11 17
22 28 33
39 44 50
80
90309
90363
90417
90472
90526
90580
90634
90687
90741
90795
5 11 16
22 27 32
38 43 49
81
90848
90902
90956
91009
91062
91116
91169
91222
91275
91328
5 11 16
21 27 32
37 42 48
82
91381
91434
91487
91540
91593
91645
91698
91751
91803
91855
5 11 16
21 27 32
37 42 48
83
91908
91960
92012
92064
92117
02169
92221
92273
92324
92376
5 10 16
21 26 31
36 42 47
84
92428
92480
92531
92583
92634
92686
92737
92788
93840
92891
5 10 15
20 26 31
36 41 46
85
92942
92993
93044
93095
93146
93197
93247
93298
93349
93399
6 10 15
20 26 31
36 41 46
86
93450
93500
93551
93601
93651
93702
93752
93802
93852
93902
5 10 15
20 25 30
35 40 45
87
93952
94002
94052
94101
94151
94201
94250
94300
'M.m
94399
5 10 15
20 25 30
35 40 45
88
94448
94498
94547
94596
94645
94694
94743
94792
04841
94890
5 10 15
20 25 29
34 39 44
89
94939
94988
95036
95085
95134
95182
95231
95279
95328
95376
5 10 15
34 39 44
90
95424
95472
95521
95569
95617
95665
95713
95761
95809
95856
5 10 14
19 24 29
34 38 43
91
95904
05062
95999
96047
96142
96190
96237
96284
96332
5 9 14
19 24 28
33 38 42
92
B637S
96421
96473
06630
96567
96614
96661
96708
96755
96802
5 9 14
19 24 28
33 38 42
93
9884S
96V.I.
96942
06988
97035
97081
07128
97174
97220
5 9 14
18 23 28
32 38 42
94
97313
97351
97405
97451
07497
97543
97589
97635
07681
97727
5 9 14
18 23 28
32 37 42
95
97772
9781*
97864
9790S
979.').".
98000
98046
98091
98182
5 9 14
18 23 27
32 36 41
96
98227
98275
98318
08408
08408
98543
98588
5 9 14
18 23 27
32 36 41
97
08677
08725
08767
98811
•
98900
08046
08089
99078
4 9 13
is 22 27
31 36 40
98
0012;
00167
00211
00251
9
09344
0038J
00432
99520
4 9 13
18 22 2i
31 35 40
99
09664
0060;
99693
09739
097 S3
00831
99870
99957
4 9 13
17 22 21
31 35 39
ANTILOGARITHMS
321
Antilogarithms.
2
3
4
5
6
7
8
9
Mean Differences
1
1
2 3
4 5 6
7 8 9
.00
10000
10023
10046
10089
10093
10116
10139
10162
10186
10209
2
5 7
9 12 14
16 19 21
.01
10233
10257
10280
10304
10328
10351
10375
10399
10423
10447
_)
5 7
10 12 14
17 19 21
.02
10471
10495
10520
10544
10568
10593
10617
10641
10666
10691
2
5 7
10 12 15
17 20 22
.03
10715
10740
10765
10789
10814
10839
10864
10889
10914
10940
3
5 8
10 13 15
18 20 23
.04
10965
10990
11015
11041
11066
11092
11117
11143
11169
11194
3
5 8
10 13 15
18 20 23
.05
11220
11246
11272
11298
11324
11350
11376
11402
11429
11455
3
5 8
11 13 16
18 21 24
.06
11482
1150s
11535
11561
11588
11614
11641
11668
11695
11722
3
5 8
11 13 16
19 21 24
.07
11749
11776
11803
11830
11858
11885
11912
11940
11967
11995
3
5 8
11 14 16
19 22 25
.08
12023
12050
12078
12106
12134
12162
12190
]221s
12246
12274
3
6 8
11 14 17
20 22 25
.09
12303
12331
12359
12388
12417
12445
12474
12503
12531
12560
a
6 9
11 14 17
20 23 26
.10
12589
12618
12647
12677
12706
12735
12764
12794
12823
12853
3
6 9
12 15 18
21 24 26
.11
12882
12912
12942
12972
13002
13032
13062
13092
13122
13152
3
6 9
12 15 18
21 24 27
.12
13183
13213
13243
13274
13305
13335
13366
13397
13428
13459
3
6 9
12 15 18
21 25 28
.13
13490
13521
13552
13683
13614
13646
13677
13709
13740
13772
3
6 9
13 16 19
22 25 28
.14
13804
13836
13868
13900
13932
13964
13996
14028
14060
14093
3
6 10
13 16 19
22 26 29
.15
14125
14158
14191
14223
14256
14289
14322
14355
14388
14421
3
7 10
13 16 20
23 26 30
.16
14454
14488
14521
14555
14588
14622
14655
14689
14723
14757
3
7 10
13 17 20
24 27 30
.17
14791
14825
14859
14894
14928
14962
14997
15031
15066
15101
3
7 10
14 17 21
24 28 31
.18
15136
15171
15205
15241
15276
15311
15346
15382
15417
15453
4
7 11
14 18 21
25 28 32
.19
15488
15524
15560
15596
15631
15668
15704
15740
15776
15812
4
7 11
14 18 22
25 29 32
.20
15849
15885
15922
15959
15996
16032
16069
16106
16144
16181
i
7 11
15 18 22
26 30 33
.21
16218
16255
16293
16331
16368
16406
16444
16482
16520
16558
4
8 11
15 19 23
26 30 34
.22
16596
16634
16672
16711
16749
16788
16827
16866
16904
16943
4
8 12
15 19 23
27 31 35
.23
16982
17022
17061
17100
17140
17179
17219
17258
17298
17338
4
8 12
16 20 24
28 32 36
.24
17378
17418
17458
17498
17579
17620
17660
17701
17742
4
8 12
16 20 24
28 32 36
.25
17783
17824
17865
17906
17947
17989
18030
18072
18113
18155
4
8 12
17 21 25
29 33 37
.26
18197
18239
18281
18323
18365
18408
18450
18493
18535
18578
4
8 13
17 21 25
30 34 38
.27
18621
18664
18707
18750
18793
18836
18880
18923
18967
19011
4
9 13
17 22 26
30 35 39
.28
19055
1909!
19143
19187
19231
19275
19320
19364
19409
194.54
4
9 13
18 22 26
31 35 40
.29
19498
19543
19588
19634
19679
19724
19770
19815
19861
19907
5
9 14
18 23 27
32 36 41
.30
19953
19999
20045
20091
20137
20184
20230
20277
20324
20370
5
9 14
19 23 28
32 37 42
.31
20417
20464
20512
20559
20606
20654
20701
20749
20797
20845
5
10 14
19 24 29
33 38 43
.32
20893
20941
20989
21038
21086
21135
21184
21232
21281
21330
fi
10 15
19 24 29
34 39 44
.33
21380
2142S
21478
21528
21577
21627
21677
21727
21777
21827
5
10 15
20 25 30
35 40 45
.34
21878
21928
21979
22029
22080
22131
22182
22233
22284
22336
5
10 15
20 25 31
36 41 46
.35
22387
22439
22491
22542
22594
22646
22699
22751
22803
22856
■>
10 16
21 26 31
37 42 47
.36
22909
22961
23014
23067
23121
23174
23227
23281
23336
23388
5
11 16
21 27 32
37 43 48
.37
23442
23496
23550
23605
23659
23714
23768
23878
23933
5
11 16
22 27 33
38 44 49
.38
23WS8
24044
24099
24155
24210
24266
24322
24378
24434
24491
6
11 17
22 28 34
39 45 50
.39
24547
24604
24660
24717
24774
24831
24889
24946
25003
25061
6 11 17
23 29 34
40 46 51
.40
25119
25177
25236
25293
25351
25410
25468
25527
25586
25645
6 12 18
23 29 35
41 47 53
.41
25704
25763
25823
25882
25942
26002
26062
26122
26182
26242
6
12 18
24 30 36
42 48 54
.42
26303
26363
26424
264&
26546
26607
26669
26730
26792
26853
6
12 18
24 31 37
43 49 55
.43
26915
26977
27040
27102
27164
27227
27290
27353
27416
27479
6 13 19
25 31 38
44 50 56
.44
27542
27606
27669
27733
27797
27861
27925
27990
28054
28119
6 13 19
26 32 39
45 51 58
.45
28184
28249
28314
28379
28445
28510
28576
28642
28708
28774
7 13 20
26 33 39
46 52 59
.46
28840
28907
28973
29040
29107
29174
29242
29309
29376
29444
7 13 20
27 34 40
47 54 60
.47
29512
29580
29648
29717
29785
29854
29923
29992
30061
30130
7 14 21
28 34 41
48 55 62
.48
30200
30259
30339
30409
30479
30549
30620
30691
30761
30832
7
14 21
28 35 42
49 56 63
.49
30903
30974
31046
31117
31189
31261
31333
31405
31477
31550
7
14 22
29 36 43
50 58 65
322
MATHEMATICS FOR TECHNICAL SCHOOLS
Antilogarithms.
Meu
Differences
1
2
3
4
5
6
7
8
9
1 2 3
4
5
6
7
8 9
.50
31623
31696
31769
31842
31916
31989
32063
32137
32211
32285
7 15 22
29
37
44
52
59 66
.51
32358
32434
3250!
32584
3265!
32735
3280!
32SS5
32001
33037
8 15 23
30
38
45
53
60 68
.52
33113
33189
33266
33343
33420
33497
33574
33051
33729
33806
8 15 23
31
30
4(
54
62 69
.53
33884
3301)3
34041
34110
34198
34277
34356
34435
34514
34594
8 16 24
32
40
47
55
63 71
.54
34074
34754
34834
34914
34995
35075
35156
35237
35318
35400
8 16 24
32
40
48
56
65 73
.55
35481
35563
35645
35727
35810
35892
35975
36058
36141
36224
8 16 25
33
41
51
58
66 74
.56
36308
36392
36475
30559
36044
36728
36813
30898
36983
37068
8 17 25
34
42
51
50
68 76
.57
37154
37239
37325
37411
37497
37584
37070
37757
37844
37931
9 17 26
35
43
52
61
69 78
.58
38019
38107
38194
38282
38371
38459
38548
38637
38726
38815
9 18 27
35
44
53
62
71 80
.59
38905
38994
39084
39174
39264
39355
39446
39537
39628
39719
9 18 27
36
45
M
63
72 82
.60
39811
39902
39994
40087
40179
40272
40365
40458
40551
40644
9 19 28
37
46
56
65
74 83
.61
40738
40832
40920
41020
41115
41210
41305
41400
41495
41591
9 19 28
38
47
57
66
76 85
.62
41687
41783
41879
41970
42073
42170
42267
42364
42462
42560
10 19 29
39
40
58
68
78 87
.63
42658
42756
42855
42954
43053
43152
43251
43351
43451
43551
10 20 30
40
BO
60
70
80 89
.64
43652
43752
43853
43954
44055
44157
44259
44361
44463
44566
10 20 30
41
51
61
71
81 91
.65
44668
44771
44875
44978
45082
45186
45290
45394
45499
45604
10 21 31
42
52
02
73
83 94
.66
45709
45814
45920
40020
46132
40238
46345
46452
46559
46666
11 21 32
43
53
64
75
85 . 96
.67
46774
46881
46989
47008
47200
47315
47424
47534
47643
47753
11 22 33
44
54
65
76
87 98
.68
47863
47973
48084
48195
48300
48417
48529
48641
48753
48865
11 22 33
45
56
67
78
89 100
.69
48978
49091
49204
49317
49431
49545
49059
49774
49888
50003
11 23 34
46
57
68
80
91 103
.70
50119
50234
50350
50466
50582
50699
50816
50933
51050
51168
12 23 35
47
58
70
82
93 105
.71
51286
51404
51523
51642
51701
51880
52000
5211!!
52240
52360
12 24 36
48
60
72
84
96 108
.72
52481
52602
52723
52845
52000
53088
53211
53333
53456
53580
12 24 37
40
61
73
85
98 110
.73
53703
53S27
53951
54075
54200
54325
54450
54570
54702
54828
13 25 38
n
63
75
88
100 113
.74
54954
55081
55208
55336
55403
55590
55719
55847
55976
56105
13 26 38
51
64
77
90 102 115
.75
56234
56364
56494
56624
56754
56885
57016
57148
57280
57412
13 26 39
52
66
70
92 105 118
.76
57.544
57677
57810
57943
58070
58210
58345
58470
58614
58749
13 27 40
54
67
80
94
107 121
.77
58884
50020
59150
59203
50420
50560
59704
50841
59979
00117
14 27 41
:>:>
69
82
96 110 123
.78
60256
60395
00534
60674
60S14
00954
61004
01235
61376
61518
14 28 42
56
70
84
98
112 126
.79
61659
61802
01944
62087
62230
62373
62517
62061
62806
62951
14 29 43
58
72
86
101
115 130
.80
63090
63241
63387
63533
63680
63S26
63973
64121
0420!)
64417
15 29 44
50
74
ss
103 118 132
.81
Ii4565
64714
04803
65013
65163
65313
05404
65615
05706
65917
15 30 45
60
75
00
105
120 135
.82
66069
66222
06374
66527
00081
60834
00988
67143
67298
67453
15 31 46
62
77
01'
108
123 130
.83
67608
67764
67020
68077
08234
68301
686«i)
68707
68865
69024
16 32 47
03
70
95
110 126 142
.84
69183
09343
69503
69663
69823
69984
70140
70307
70469
70632
10 39 48
64
81
07
113
129 145
.85
70795
70958
71121
71285
71450
71614
71770
71945
72111
72277
17 33 50
(,^
83
89
110
132 149
.86
72444
72611
72778
72046
73114
73282
73451
73621
73961
17 34 51
68
86
101
118
135 152
.87
74131
74302
74473
74645
74S17
74989
75162
75336
75500
75683
17 35 52
69
87
104
121
138 156
88
75858
76033
76208
76384
76560
76736
76013
770!K)
77268
77446
71
80
107
125
142 150
.89
77625
77804
77983
78163
78343
78524
78705
78886
79068
70250
18 36 54
72
01
100
127
145 163
.90
79433
79616
79799
79983
80168
80353
80538
S0724
80910
81096
19 37 56
74
93
111
130
148 167
.91
81283
81470
81658
81846
82035
82224
82414
82604
82794
82985
19 38 57
76
95 113
132
151 170
.92
Vi 171.
S330S
83560
83753
83046
84140
84528
81723
84918
19 39 58
78
97
116
136
155 175
93
S5114
85310
85507
85704
85901
86099
86298
86497
86696
80896
20 40 60
70
99
11!)
130
158 178
.94
87096
87297
87498
87700
87902
88105
88308
88512
88716
88920
20 41 61
81
102
122
142
162 183
.95
89125
89331
89586
89743
89950
90157
90305
90573
90782
90991
21 42 62
83
101
125
146
166 187
.96
.11201
91411
91022
91833
92045
02257
92470
92083
02897
93111
21 42 64
85
106
127
149
170 191
.97
13325
03541
93756
03072
04180
01 106
94624
94842
96060
95280
22 43 65
87
109
130
152
174 105
.98
15499
05710
05040
06161
96605
96828
97051
07275
97499 22 44 67
39
11
133
155
178 200
.99
'77:' 1
07010
98175
98401
0S628
ivs.V,
99312
0054 1
00770 23 40 68
>1
14
137
lik)
182 205
ANSWERS
Page
4.— 7. 11413. £.23437. 5.16438. 439226. 5.36509.
6. 190780. 7. 7907. 8. 115894. 9. 222065.
10. 8380882. 11. 35880920. 12. 109648526.
13. $105695. U $222139. 15. $225315.
5.— 1. 7535. £.236. 5.364947. 413920.5.3440879.
6. 3207. 7. 10330. 8. 24116. 9. 11232. 10.
217209.
7.—1. 1557. 2. 1309. 3. 6152. 4 952. 5. 55368.
6. 2960. 7. 34500. 5. 7645. 9. 7864. 10. 11800.
11.1788. 12.17880. 15.7864. 14. 0469. 15.469.
16. 42905. 17. 18408. 15. 78. 19. 7614. 00. 1180.
21. 236896000. 22. 2538. 25. 1018400. 2k. 6885.
5.— 1. 3276. 2. 6344. 5. 4796. 4 31433. 5. 67392. 6. 3213.
7. 1910520. 8. 2809566. 9. 5049668. 10. 699678.
11.49434. 12.26430588. 15.1299276. 14 69765.
15. 021112. 16. 038934. 17.2858392. 18. 7916832.
19. 50092640436. 20.. 13255716. 21. 11109280.
22. 147504. 23. 78028125. 24 2112280092.
25. 95797296. 26. 338928. 27. 29716736. 28.
1925625. 29. 0000625. 30. 155697696.
10.— 1. 32987. 2. 180912. 5. 1764. 4 11^ 5. 00158.
6. 176. 7. 01325. 8. 122. 9. 025. 10. 851.
11. 1702. 12. 31196i 13. 4642.
75.— 1. 6502. 2. 64. 5. 67965. 4 211992. 5. 13429.
6.27810. 7.304. 5.797. 9.1419. 10.407117.
11. 29799. 12. 1033.
lk.—l. 18. 2. 20. 5. 44. 4. 70. 5. 6891. 6. 585. 7. 183.
15.— 1. 28. 2. 3f. 5. 9. 4 240000. 5. 230^. 6. ^A ?• t¥t
8. 2ff 9. 8. 10. 48. 11. 768. 1& 1152. 13. 25.
14 22$. 1. 1133 lb.
323
324 MATHEMATICS FOR TECHNICAL SCHOOLS
Page
16.— £.229175. 5.252 ft. 43199. 5.3773,3507. 6.465481b.
267721b. 7.2361b. 5.144. 9. 14 rem. 2 ft. 10. 177$ miles.
11.60. 12.251 miles. 13. $1028716. 14.2400.
17.— 15. 8. 16. 72.
Of) 1 6 O 20 <? 4 5 /. 21 K 1 5 /? 4 9 7 102 O 6 8
Q 32 If) 72
*• ^5"^" • /t7  rr&"
£>/) 7 1 O X .?i /, ^ T2 /J 3 7 7 242 O 107
Q 9 rn 5
t7 ' 100  117, 16
21.— 1. 3£. £. 4. 5. 7$. 4 6. 5. 8f. 6. 17f 7. 29$. 5. 3ft.
9. 11 T 2 T . 10. 27^. 11. 2500. 12. 9&. 15. 53. 14. 9.
15. V° 16 ts I 7  ff ^ H¥ I 9  H 1  20. *$*.
£1. SJg 5  8 . flg. 1 ^2o 21  ^ w 3  H> ^g^.
22.— 1. H *• If 5. A 4 Iff 5. $ff. 6. f$. 7. 1^. 5. Iff.
9. $f. 10. 11. 11. 3ff i*. 3^fr. 15. 19ft *4 8Hf.
£5.1. 33f lb. 0. f. 5. 2^". 4, 2$f ". 5. 24 T 5 F . 6. 19$*.
£4.7. 3f*. 5. 16ft', 11*. 9. 2A', 2f , Iff. ^ If;
26.— 1. 2f. £. 2f. 5. if. 4 A 5. ft. 6. A. 7. / T . 5.^.
9. If, 10. f$. 11. M$*. 12. 1 ¥ V 15. 500. 14. 4f.
15. 62. 16. 25. 17. 3. 15. 1. 19. 5^ T . £0. 39ff. JM. 49.
22. 4f. £5. 1. 2Jf. 27ff
27.—1.&. 2. ft. 3. ^. 4. A 5.^. 6. if. 7. 3f. 8.376$.
9.6. 10.^. 11. 7ff. 12.$$. 15. 12fff. ** riV. 15. 1^.
16. 16$f
©7 7 4 3 £> 1 <? 94 /, 3567 K 251 fi 1 7 14 1
*'• l  T0T7 *• ITS' J  TZS 4 50^0 ° 507 ° T27 '■ IToo
O 617 Q 2 1/1 31
° "S~ooTT a  TT5  LU  200000
28.— 1. 25. 0. 5. 5. 375. 4 6875. 5. 6. 6. 992. 7. 9G875.
5. 9375. 9. 96. 10. 9921875. 11. 74. 1#. 508.
29.— 1. 5. 0. 083. 5. i42857. £ 13. 5. 05. 6. • is.
7. 307692.
50.1. f. 2. A *.**• WiV 5tVs StVt 7.3$f * ff$
Ql /rt 9 23 7/0589 1® 78 7 7<3 11 7/ 7 1
tf. T . iU. Z^q. 11. ^ TT xo <**• 166 5 0 IO ' 3 00 0 J ^ T¥T5"
M.— *. 4, 2, 22i 60. 5. 50%, 25%, 12f %. 4 50%, ?
33*%. 5. 416%, 4$$%. 6. 15, 35, 70, 85. 7. 80. 5. 70.
5£.— 9. $807£. 10. 30 gallons.
ANSWERS 325
Page
33.— 1. 10, 5, 10. 2. $7830. 3. 11^ hr. 4 $16747.
5. first by 4c. 6. 212 lb. 7. $7225. 5. 9ff. 9. 4.
10. $84000. 11. $169290. 1£. $312500. 13. $2563141.
r /, 4 2 1
54.— i5. ff in. 16. i 17. 8f. IS. 10 lb. 19. $2703f.
20. 73. 21. $844. &g. 5494 lb. copper, 2706 lb. zinc.
23. 9 T Vr days. #4 $36000. £5.63295. £6.309462.
27. 232 oz. £3. 6 T 6 g in.
35.— 29. 5597". 50.8327. 31.45225". 3£. 96. 33. 83f.
34. 78. 35. 25, $4000. 36. 10625, 1875. 37. 14 in.
38. 82264 cu. in. 39. 72. 40. 8000 lb. 41. 11§ in.
36.— 4£. If in. 43. $577500. 44 $410000, $615000,
$615000. 45. 107^, 35f£%. 46. 65f. 47. 1038336.
38.— 1. 1760. 0. 5280. 3. 027. 4. 003125. 5. 681".
6. 16339^'. 7. .000125. 3. 00126. 9. 415 li. 10. l^&ch.
11. 287496'. 12. ^ ch. 13. 5237 miles.
39.5. 14', 44406". 6. 9525 cm., 09525 dm. 7. 06096 Km.
40— 8. 981456 cm. 9. 134 Km. 10. 134 112 cm.
41.— 4. 4840 sq. yd. 5. 1321 sq. ft. 6. 033 ac. 7. 6930 sq. in.
3. 51 sq. yd. 9. 2} ac. 10. 803 sq. ft. 1J. ff sq. ft.
1£. 198. 13.96. 14.572V 15. 216 ac.
42.— 16. 24012. 17. $37333. 13. S653. 19. 72. 20. 54ff.
43.3. 225 cu. ft. 4 143432 cu. in. 5. 6790363 cc. 6. Iff
7. 3^oV 37. 3. 21^y, 104. 9. $1067. 10. $14344.
11. 129.
44—12. 44245. 13. 19' 6". 14 $11212, $12834. 15. \".
16. 36300.
46.— 1. 5835 cu. ft. 2. 183.
47.— 3. 103. 4 90864018 tons. 5. 79. 1.196. 2. 484 days.
3. 45563 min. 4 1496.
49.3. 343 1, 343000 g. 4 3875 cu. in., 1096625 lb". 5. 2268
Kgm. 6.52310. 7.627198. 3.1214514. 9. 3600 lb.,
°2T
51.— 1.45. £.199. 3.123. 4327. 5.37. 6.1732. 7.34908.
3.119668. 9. 7370. 10. 2507. 11.220yd. 12. 1167".
13. 3059". 14. The square pipe.
326 MATHEMATICS FOR TECHNICAL SCHOOLS
Page
52.— 15. 1587". 1. 1414". 0.4882'. 5.5138'. 4.12727'.
5. 46904'.
53—6. 12196'. 7. 2085'.
5k.— 1. 639. 0. 593. 3. (a) $15905. (6.) $14633.
55— U 22897. 57242, 114448. 5. 2809f
56.— 0. 21409. 5.30491. 4. $22020, $31363. 5.66248.
6. $264717.
59.5. 6, 72. 6. 13$, 160. 7. $14360. 8. $13978. .9. $4469.
65.— 2. $546, $1176, $2449. 5. $800, $1493, $3499.
66.— J,. $2500, $3386, $5233. 5. $9563, $11455, $49640.
6. $4712.
67.7. $6483. 1. (a) $5702, (6) $8859, (c) $19750.
0. (a) $9104, (6) $9216, (c) $17340. 3. $18021.
68.4. $37528. 5. $80042.
69.— 1. $697, $1039, $825.
70.— 2. $497, $220, $1020. 3. $1011, $899, $850.
89.— 1. 70', 14'. 0; 320. 3. 620'. £. 160. 5. 5760 sq. ft.
90.— 6. 1501". 7.120'. 8.12. 9. $600000. 10. 4800 cu. ft.,
1200 cu. ft, 11. 15'. 12. 18", 6". 13. 48c, 32c, U. 65c,
26c 15. 2 years. i6. 3 days. 17. 9 T V min. 2£. $f day.
91.— 19. 25 miles, 30 miles. 00. 16f. 0i. $200000. 22. 68 T 2 T .
05. 2100 gals. 2A. $6875000. 25. $1800000. 26. 4£'
from fulcrum. 27. 10 cm.
93.— 11. 19 T V 20. tV **• 4 !§ *•*• If
9^.— 1. 2a. 0. 2a3o3c. 5. 8x + 23?/+3z.
4. 7syyz+10zx. 5. 3z 2 + 3x6. 6. 7a. 7. 6x 3 2x 2
*+2. 8. 4# 3 + 2 / 2 + 3.
102.—l.x+±. 2. a + l. 3. a2. 4. *"7. 5. 3x4. 6. 32a.
7. 2+z. 5. s+y. 9. 53a. 10. 2x 2 + 7. ii. z+b.
12.x 2 +xy+y 2 . 13.a 2 ab+b 2 . U. x*+Zz % y+%xy*+y t .
15. a 2 \b 2 \c 2 — ab — bc — ca.
104.— 2. \, 88, 375. 3. 0315, 477, 2381. & 26775, 480, £.
5. 15, 3. 6. 8jSj, 73i 2i 55, 110.
ANSWERS 327
105.7. 26832, 244, 000327. 5. 104, 024. 9. 23184, 14.
10. 448, 254375, 2056, 1155. 11. 65625, 375,
40533, 75, 7875, 1083.
106.— 12. 442, 2238, 1865. 13. 1584, 3472, 52 09, H. 300,
20, 25. 15. 476, 3398, 34375, 40333.
107.— 16. 34f, 291§, 4, 17181.
113.— 3. 6, 181, 1394.
114.— 5. H ft. 6. 72. 7. $25800. 8. 192 sq. in. 9. 10 lb.
10. 5712. 11. $15728. 12. $40460.
115.— 13. $1080. 14. $973. 15. 7936 acres. 16. 5H sq. ft.
17. 137218, V3 half the base. 19. 23382 sq. in.
20. 1393 sq. in. 21. 15. **. 60. 05. 224. 24 111".
116.— 25. 10825 sq. in., 62352 sq. in., 73177 sq. in.
06. 875 sq. in. 27. 41 sq. in. 28. 77 sq. ch. 09.375
sq.ft. 30. 312 acres. 31. 1425 sq. ft. 50. 6605.
117.— 33. 72703. 54 34208. 55. 1777.
118.— 36. 1581. 57. 149', 94', 828^ sq, ft. 55. 8912 sq. ft.
105.— 2. 131 95". 5.471'. 44618,5248. 5.599". 6.240.
7. 376992. 5. 188496. 9. 1528". 10. 180418 miles.
11. 14". 10. 140, 360. 13. 300 R.P.M. 14 13".
124.— 18. 31416, 64403 lb. 19. 963 sq. in.
125.— 20. The 4" pipe. 01. 4, 9. 00. 5383". 05. 1236".
24. 156". 06. 1492 sq. in. 07. 87965 sq. in.
05. 638727 sq. yd. 29. 3456 sq. in. 30. 761 sq. in.
51. $941491. 50. 2011 sq. ft.
126.— 33. 187". 54. 525 sq. in. 35. 4133 sq. ft. 36. 672
sq. in.
132.— 3. 54978', 23562 sq. ft. 4 4411 lb. 5. 22". 6. 28".
7. 649 sq. in. 8. 1325.
133.— 9. 12311 sq. ft. 10. 373 lb. 11. 10454'. 10. 866
. sq.ft. 16. 598113 sq. in. 17. 202 1 sq. ft,
157.1.4:3. 0.5:2. 5.3:5.
155.4.168,112. 5.335:6.6.20:49. 7. 196 T 4 T . 5.175. 9. $144.
10. $5625. 11. llf 10. $5334. 15. 833'. 14. 51f.
15. 54'. 16. 124, 93, 31. 17. 60, 24, 12.
328 MATHEMATICS FOR TECHNICAL SCHOOLS
Page
139.18.4^'. 19. 28£". 20: ^^ 21. 15'. 22. 11,80'. 28. 1056'.
U3.—1. 1200, 800. 2. \ ,  5 T °. 3. 66§, 133J, J. 5, lOf.
5. T \, 5, P= T VW+5. 6.6000,20. 7. 046, 5. 5.216,
•0044. 9. 1957, 1521. 10. 345, 245.
1U.—11. 100, 200. 12. 4&, 5J. 13. 10615, 7221. U. 1042,
•0088. 15. $21000000, $1890000. 1. 135, 326, 32.
2. 94, 15.
U5.—3. 102, 210, 213, 14. *. 11904, 20, 10. 5. 548, 106f,
63, 21504, 658. 6. 30062 1, 377, 65.
146.— 7. 134, 122, 7, 16. 8. 6283200, 31416, 079. 9. 58,
1125, 7. 10. 172, 1616, 248. 11. 1112, 217,
•06,666.
U7.—12. 171", 445". 13. 24525", 315". 14 666, 10692.
15. 186", 349, 4400. 16. 11", 5818, 3300.
W.— 17. 42 215, 35 9, 6 9. 18. 10472, 143, 86450. 19.6676,
18390, 14921.
W.— £0. 3, 11819.
179.— 1. 1964. 0. 6875.
180.— 3. 6363. 4 39285. 5. 2292. 6. 955". 7. 2386.
8. 1145". 9. 2357 + . 10. 7636.
181.— 1. 2304. £. 960 min. 3. 16 min. 4 1131 min.
5. #f. 6. 5 min. 7. 0028". S. 5 min. 9. 792.
10. 3394.
187.— 1. ". 2. 1". 3. 8625". 4 12". 5. 5". 6, 7^".
7. No. Morse. 8. B. & S. 9. 604". 20. No. Mq
11. \". 12. Jarno. 14 if". 15. £". 16. ^" . 17. T %".
188.18. (a) f", (6) T y, (c) \", (<£) f ". 1.9. If. 50. 4".
£1. 3^", tV'. 22. 21". 23. 168". ££. 2° .V.)'.
2° 52' 24", 2° 23', 2° 51' 50".
191.— 2. 12. 5. 10. 4 H 5 H «• H 7 *• *■ ; '
9.1". 10.1615".
193— 1. 1625". 5. 4541". 3. A 4 4". 5. A 6. 4.
7. 3. S. U. 9. *. 10. *
ANSWERS 329
Page
195.— 1. 1187". 2. 2167". 5. . 4 3". 6. If 7. .
5. 4. 9. 6.
i97._i. 05336", 01144". 2. fr, 62193". 5. 08004",
•83992. 4 0915", \, 0196". 5. 4£, 1 7154", 02948".
6. 04", 295", 00858".
202.— 1. 24, 64. 0. 32, 48. 3. 45. 4 16. 5. 24, 72. 6. 36.
7. i 5. 24. 9. 72. 10. 42, 98.
205.— 1. 4\. 0. 24 stud, 92 lead, 36 inside c, 72 outside c.
206.— 3. 48 stud, 28 lead, 36 inside c, 72 outside c. 4 24 stud,
96 lead, 72 inside c, 36 outside c. 5. 42 stud, 98 lead.
6. 24 stud, 96 lead. 7. 64 stud, 40 lead. 5. 112. 9. Equal
gears. 10. 36 inside, 72 outside.
210.— 1. 8". 2. 2618". 3. 1122". 4 1541". 5. 0982".
6. 1348". 7.5166". 5.55". 9.70. 10.46. 11.10.
12. 7".
212.1.4:19. 2.40. 5.409". 4.42. 5.304. 6.509".
7. f" per min. 5. 2" per min.
213.— 9. 5. 10. i".
225.2. 15". 4 15". 5. 21333". 6. 215". 7. 49° 57'.
226.8. 22° 48'. 9. 1". 10. 14945". 1. J». 2. £. 3. 10.
4. 1443". 5. 9. 6. 3466", *. 7. 9f min. 5. 7l£.
9. \", 2° 23'. 10. ". 11. 5487". 12. T V 13. 28.
227.— 14 8. 15.72. 16.36. 17. llj. 18. 5. 19. 36 inside,
72 outside. 20. 65". 21. 11". 08. 382". 25. 6f.
2k 24 worm, 32 second stud, 64 first stud, 72 screw.
25. 33° 8'.
239.— 1. 30099. 2. 4420. 5. 08836. 4 89827. 5.75229.
6. 21015. 7. 33332. 5. 58798. 9. 0055873.
10. 0007237. 11. 58009. 12. 44419. 13. 12057.
14.1878. 15. 2999. 16. 25507. 17.237. 15.180.
19. 168.
240.— 20. 27904 sq. yd. 21. 18886 sq. in. 22. 43301
acres. 25.13221. 24.4754'. 25.5077". 26.16495
sq.in. 27.3073. 25.13365. 29.40101 + . 50.59221.
330 MATHEMATICS FOR TECHNICAL SCHOOLS
Page
242.2. $2720. 5. 6835. 4 472. 5. 1057 sq. ft.,
229 cu. ft. 6. 541.
2U—3. 12818 lb. 4. 11846+. 5. $2618.
245.6. 28 • 49 lb. 2. 502 • 65 sq. in.
246.3. 667 • 98 lb. 4 2127 • 89 lb. 5. 522 • 78 lb. 6. • 280 lb.
248.2. 48 sq. in., 1146 sq. in. 3. 5003 lb. 4 1615 1b.
5. (a) $7197, (b) 22200 + .
249.6. 7086.
250.— 2. 31196 lb. 5. 1314 sq. in., 2411 lb. 4 78564,
$4974. 5. 359 cu. in.
254.— ». 91 sq. in. 3. 49661 lb. 4 4801 lb. 5. 3073 cu. in.
6. 236 cu. in.
256.3. 6702. 4 4134 lb. 5. 155 lb. 6. 229 lb.
257.— 1. 907 sq. ft. fc 6113 sq. ft.
258.3. 7422 sq. ft., 4186 cu. ft. 4 35219 sq. in., 49009
cu. in. 5. 83776 cu. in.
259.— 1. 30 1b. 2. 2988 tons. 3. 332564 lb. 4 7854 1b.
5. 16593. 6. 10102+ lb. 1. 141". 2. 8216 lb.
260.— 3. 5 12". 4 153/ 5.1653'. 6.2 02 1b. 7. 7068 lb.
8. 2501'. 9. 329148 sq. ft. 10. 484. 11. 2253.
12. 5130 sq. in. 13. 20 min. 14. 13675+ lb.
15. 603 19 sq. in. 16. $1700. 17. 16317.
261.— 18. 2579. 19. 2787 cu. in. 20. $25500. 21. 2651 lb.
22. 877430 lb. 23. 243. 24 265". £5. 2215 lb.
26. 501". 07. 11581 + . 28. 648000. 09. 1942 1 sq.
ft., 10603+ cu. ft. 30. $7367.
262.— 31. 1257 cu. in. 50.332". 55. 19215+ lb. 5^. 7979.
55. 376 99 sq.m. 36. 4 69 lb. 57.1450". 55.1376.
39. 1284". 40. 959".
264.— 11. (ab)(xy). 12. (x + z)(xy). 13. (3 + a)(xy).
14 0r 2 ?/)(.T2). 15. (ax b) (bx a). 16. (a 4 + l)(a + l).
17. (a 2 6)(l+c). 15. (a +3) (0a 2  c). 19. (a: + m 2 )(x+m).
00. (a + b){x + yz).
ANSWERS 331
Page
267.5. (a+b + c)(abc). 6. {x 2 +y 2 ){x+y){xy).
7. (x*+y*)(x 2 + y 2 )(x + y)(xy). 8. (a + b + c)(abc).
9. (x + y + a + b)(x + yab). 10. (x + y)(x+y)(xy)
(xy).
268.— 11. (x+yz)(xy + z). 12. (la6)(l + a + 6).
13. (a 8 + l)(a 4 + l)(a 2 + l)(a + l)(al).
Ik. (xy + a + b)(xyab). 17. 5(a6+2c)
\ab2c). 18. (4a + 6)(4 + a6).
19. (2x 2 +xy+3y 2 )(2x 2 xy+3y 2 ).
20. (x 2 3x + 9){x 2 +3x + 9).
21. (x 2 + 2xy + 2y 2 )(x 2 2xy+2y 2 ).
22. (x 2 xl)(x 2 +xl)(x i +Sx 2 + l).
23. (2x + y)(2xy)(x3y)(x+3y).
2k. (x 2 + 2x+4)(^ 2 2x + 4).
25. (3xy)(3x + y)(x + y)(xy).
26. (2x + 3y)(2x3y)(x+y)(xy).
27. (x 2 xy + 3y 2 )(x 2 +xy + 3y 2 ).
28. (x 2 3x+5)(x 2 +3x+5).
269.3. (x 2 +l)(x 4 x 2 + l). k. (ab)(a+b)(a 2 +ab+b 2 )
(a 2 a& + 6 2 ). 5. (x + 2)0r2)O 2 +2z+4)(x 2 2z+4).
6. (a6)(a 2 + 6a+36). 7. 3(1 3x)(l+3x + 9:r 2 ):
8. x(x3)(x 2 +3x + 9). 9. 2(:r+5)0 2 5x + 25),
10. (xy)(x 2 +xy+y 2 )(x + y)(x 2 xy+y 2 )(x 2 + y 2 )
{x 4 x 2 y 2 \y i ). 11. (a + &c)(a 2 +2a& + 6 2 + ac + &c + c 2 ).
12. 2b(3a 2 + b 2 ).
272.— 1. i :. 2. .. 3. x 3 . A. 18. 5. f. 6. 1. 7. 1. S. a 2 .
a; 2 p 6 2
9. \. 10. 4i. 11. 1. 12. a 3 .
a 3 2 a
27*— 1 * 2a 3  k — 5 £ 6  7 
ax
8. 2xK 9. 7. 10. —,. 11. x. 12. 2yK 13. — L .
a* xi x*
14. ^. 15. ^. 16. 8. 17. . 15. f . 19. 64. 00. 8.
«i. ^. ^. si 23. 8. 24 i ^5. 8. 26. 5. 27. 2.
28. f. 29qr 4 .. 50. W «■ ¥• **•***• 56.^/19863:
332 MATHEMATICS FOR TECHNICAL SCHOOLS
Page
273.37. ^2187. 38. #9. 39. xy. 40. xi+xyWyyl
41. 8. 42. T V 43. C V U 81.
275.— 1. ^P, ^4* \76l 2. a^/5 1 , ^TT 2 , ^13. 3. #2*,
V2», V2\ 4. V8», v^ $'6«. 5. V6\ 6. V30.
7. V¥. *. \Z4500. 9. 2^/9.
276.— 1. 25 V7. £. 8\ZlT. 5. 14 V^ . 5V5". 5. 1697.
6.1212. 7.1581. 8.3674. 9.7826. 70.3175.
ii. 7. _2£. 265. 25. 2598. U. 735^ 15. 1626.
16.4^4. 17.6^2. 18.5^/5. 19.9^3. £0.3429.
£1.96. #& 6124, 8$. 4157. *£ 11878. 05.5196.
00. 33255.
278.— 1. 866. 0. 1155. 5. 9794. 4 2683. 5. 204.
6. 19596. 7. 403. 8. 045. 9. 257. 19. 315.
11.1702. 12.822. 15.8989. 14. 809. 15.1869.
16.2184. 17. 101. 18.1768. 19.1294.
20. 46647.
282.— 1. V, 5. 2. ¥, V 3. f, 3. 4. 2, 1.
5. 3, 8. 6. 3£, 1. 7. 172, 128. 8. 619, 807.
9. 238, 462. 10. T % f. 12. f, f. 12. 3, 1.
13. 2, . 14. 13, f. 15. 12, 2. 16'. 3, f. 17. 4, f
2S.f^. &=£•=£. 20.3a,f.
285.— 1. 5, f. 2. i 9. 5. i 2. 4 2, 172. 5. f, 5.
6. 3, 2. 7. 1, §. 3. ±3, ±2. 9. &, f.
10. ±2, 5, 1. 11. ±3, ±4. 12. 3, 2. 13. 2\. 4.
14. 6", 8". 15. 5". 16. 323 rods. 17. 3 82", 6 18".
18. 437 sec. 19. 759 sec.
287.— 1. x = 17, 11, y—11, 17. 2. x = 14, 9, y = 9, 14.
8. x = 27, 19, y = 19, 27. 4. * = 71, 13, ?/ = 13, 71.
5. x = ±5, y = ±7, x = ±7, y = ±5. 6. x = ±8, ±5, ?/
= ±5, ±8. 7. x = 13, 3, i/ = 3, 13. 8. x = ll, 8,
y = 8, 11. 9. * = 1, 2/ = l. 10. 4", 3". 11. 5", 7".
20. 24", 7".
ANSWERS 333
Page
290.— 1. 20 tons. 2. 1817 lb. 3. 480 ft. per sec.
4 126f ft. per min.
291.— 5. f". 6. 997 lb. 7. 20 2". 5. 2089 lb.
9. 23361+ #.P. m 1055". 11. 2406". 12. 285".
29k.— 1. 1191'. «. 1136'. 3. 2539 miles per hr. £ 9.
295.5. 37. 6. 307 ohms. 7. 2X3 3 = 1 X(378) 3 approx.
5. 15 lb. 9. 2'. 10. 9 sq. ft. 11. 51 ,66 cc.
296.— 12. 2365° C. 13. 324 ohms.
300.— 1. $49205. 2. $1000. 5. $1000. 4 $115060.
5. $220805. 6. $479496, 6% interest.
301.— 7. 8576104. 8. $112854. 9. $368040, $205500.
10. $741340. 11. $10,13270. 12. $22803, $68398.
13. Amount of premiums invested = $7325 • 60.
301.— 1. 3*. 2. 310.
302.— 3. 33. k $159120. 5. 615^. 6. 228, 40, 60.
7. $11491. 8. lf$ hr. 9. 21 min. 10. 16.
11. 31174 lb. 12. 80. 13. 286. 14 *. 15. 444
hr. 16. 833.
303.— 17. $ll27i 18. 539. 19. 4526. 20. 224^, 492.
01. 406. 22. 389 sec. &f. $399. 2k. 62£.
05. $12267. 26. 65%. 07. 2925 lb. 28. 104.
09. 899. 50. 112, 6.
30^.— 31. $5700. 50. 41f, 33i 25. 33. $3199. 34. f.
35. 28 • 28'. 36. $210 • 32. 37. 19 lb. 55. fj. 39. 8 • 67".
40. 18113 lb. hi. 144317 lb. 40. 86, 31, 312.
45.280. U. 3697 lb.
305.— k5. 42 + . 46. 5027 sq. ft., 7697 sq. ft. 47. 40'.
48. 83". 49. 1273". 50. 942". 51. 3278 sq. in.
50. 776. 53. 1146, 10". 54. 1326'. 55. 532".
56. 125". 57. 1016 lb. 55. 215. 59. 838.
306.— 60. 147". 61. 501". 60. 7i". 65. 796".
64. 122140. 65. 1575". 66. 4826 lb. 67. 343.
65. 1333i 640, 213i 69. $85271. 70. $38963.
334 MATHEMATICS FOR TECHNICAL SCHOOLS
Page
307.— 71. 92", 253". 72. $4256. 73. 491. 74. 8183.
75. 235 + . 76. 11875", 118 06", 58. 77. $50752.
308.— 78. 22135. 79. 141372. 80. $52393. 81. 3 08.
82. 5887. 83. 18272. 84. 102 lb. 85. 19867.
86. 5877, 397 sq. ft. 87. 67259 sq. in., 191 lb.
309.— 88. 1006, 631. 89. 17413 cu. in. 90. 14766.
91. $24183. 92. $10971. 93. 11103. 9k 5331.
95. 2575 • 7 lb. 96. 148 • 68", 147 • 92", • 52. 97. $234 • 13.
310.— 98. $14272. 99. 32 R.P.M. 100. 66159 sq. in.
311.— 3. 2", 3", 1". 4. 615. 5. 52° 30', 37° 30'. 6. 18955
lb. 7. 528". 8. 75000. 9. 435". 10. 72°, 72°, 36°.
ii. 81. 12. 16", 11314". 15. 503. lit. 648".
15. 649. 16. 1549".
312—17. 592. IS. 36'. 19. 577", 1154". £0. 155938.
21. Sides 8, 12, perp. 10. 22. 23395 X10 4 . 23. 325
miles. 24. 47% too great. £5. 3", 4". £6. 6".
27. 26966 sq. ft. 0$. 13", 15". 29. 40'.
313.30. 28". 91. 64'. 30. 1327'. 3k 6"X12". 35. Z7\.
36. 292". 37. $90000. 38. 11652 oz. 39. 9", 12",
15". 40. 4".
314.— 41 3600. ^. 0039.
INDEX
Abscissa, 150.
Addendum, 209.
Addition, in Arithmetic, 3 ; in
Algebra, 92.
Annulus, 121.
Antilogarithm, 231 ; tables, 321.
Ashlar, 54.
Axes of reference, 150.
B
Bead, volume of, 257.
Board foot, 56.
Brackets, 78.
Brick work, 55.
c
Calipers, outside, inside, herma
phrodite, 171 ; vernier, 177.
Cancellation in Arithmetic, 14.
Centring, 172 ; by hermaphrodite
calipers, 173 ; by centre square,
173.
Characteristic, 229.
Circle, 118 ; circumference, of,
119 ; area of, 120 ; arc of, 121 ;
sector of, 122 : segment of, 122.
Circular ring, 121.
Clearance, 209.
Coefficient, 71.
Cone, 246, 247.
Coordinates, 150.
Cylinder, 243 ; hollow, 245.
Decimal point, 1.
Decimal fractions, 27 ; repeat
ing, 28.
Decimal equivalents, 315.
Decorating, 68.
Dedendum, 209.
Division, in Arithmetic, 3, 9 ; in
Algebra, 100 ; rule of signs, 100.
Eaves, 61.
Ellipse, 126 ; to construct, 127 ;
area of, 127 ; circumference of,
127.
English linear measure, 37.
Equation, simple, 84 ; simul
taneous, 141 ; quadratic, 279 ;
simultaneous quadratic, 286;
exponential, 238.
Exponent, 74.
Expression, Algebraic, 71.
Factors, in Arithmetic, 14 ; in
Algebra, 263.
Feed, lathe cutting of, 180 ; of
milling machine, 211.
Field book, 112.
Flooring, 57.
Formulas, 103, 144.
Fractions, definition of, 18 ; kinds
of, 18 ; addition of, 21 ; sub
traction of, 21 ; multiplication
of, 25 ; division of, 26 ; deci
mal, 27. *
Frustum, 251, 252.
G
Gear trains, 197, 198 ; compound,
199.
335
336
MATHEMATICS FOR TECHNICAL SCHOOLS
Gears, calculation, 207 ; reduc
tion in headstock, 205 ; quick
change, 206, 207.
Geometrical series, 297, 298.
Graphs, 150.
Guage of slate, 65.
Heel, 61.
Headstock, 205.
Imaginary quantity, 284.
Index, 74 ; laws, 75, 270.
Index plate, 213, 215, 219.
Indexing, rapid, 213 ; plain, 214
differential, 217.
Irrational quantity, 274.
Irregular figures, area of, ]29.
Lathe, cutting speed of, 179;
compound geared, 202; 203;
lead of, 201 ; simple geared
200.
Lathing, 57.
Lead screw, 200.
Logarithm, 228, 229 ; of number
less than unity, 235; of a
power, 236; tables, 319.
Lumber, 56.
M
Machinist's scale, 171.
Mantissa, 229. •
•Measure, linear, English, 37;
linear metric, 38 ; square,
English, 40 ; square, metric,
40; cubic, English, 42; pubic,
metric, 42.
Micrometer, 175.
Milling machine, 210; cutting
speed of, 210; feed of, 211, 212 ;
lead of, 223; change gear cal
culation, 223.
Multiple, least common, 22.
Multiplication, in Arithmetic, 3,
5; tables, 6; in Algebra, 94';
rule of signs, 95.
N
Negative quantities, 81.
Notation, in Algebra, 71 ; i n
Arithmetic, 1.
Ordinate, 150.
Origin, 150.
Outside diameter, 209.
Painting, 68, 69.
Parallelogram, 109.
Percentage, 30.
Pictograph, 162.
Pitch, of roof, 61.
Pitch, diameter, 208 ; circle, 208 ;
diametral, 208 ; circular, 208.
Planimeter, 130, 131, 132.
Plate, 61.
Plastering, 67.
Polygon, area of, 127.
Power of, 10, 7 ; ofaquantity, 71.
Present worth. 299.
Prism, 241, 212.
Prismoid, 259.
Proportion, 135; inverse, 138; in
similar triangles, 136.
Pyramid, 249.
n, value of, lift,
Q
Quadratic equations, 286.
INDEX
337
R
Rafters, 01; hip, 62, 63; jack,
62, 63.
Ratio, 134.
Rational quantity, 274.
Rectangle, 108.
Ring, solid, 258 ; anchor, 258.
Rise, 61.
Roofs, gable, 61 ; hip, cottage, 62.
Roofing, 64.
Root, square, 53 ; in Algebra, 76.
Run, 61.
Rubble, 54.
Screw, 188.
Shingles, 64.
Signs of operation in Arith
metic, 13.
Simpson's Rule, 129.
Simultaneous equations, 141 ;
simultaneous quadratics, 286.
Slate, 64.
Span, 61.
Specific gravity, 46 ; tables of,
316, 317, 318.*
Sphere, 254, 255 : sector of, 257 ;
segment of, 256 ; zone of, 257.
Spirals, cutting, 221 ; position of
table, 224.
Square, 108.
Square root, 50.
Stone work, table, 54.
Subtraction, in Arithmetic, 3, 5 ;
in Algebra, 93.
Surds, 273; quadratic, 274; like
and unlike, 274 ; addition of,
275 ; subtraction of, 275 ; mul
tiplication of, 275 ; mixed and
entire, 276 ; division of, 277.
Symbols, of Arithmetic, 1 ; of
Algebra, 71.
Taper, as amount, 183 ; as angle,
183; Morse, 183; B. & S„ 184;
Jarno, 184 ; cutting by com
pound rest, 184, 185 ; cutting
by offsetting tailstock, 184, 185,
186 ; cutting by taper attach
ment, 186.
Terms, like and unlike, 77.
Threads, pitch, 188; diameter of,
188, 209 ; inside diameter of,
188, 209 ; single, double, triple,
189 ; righthanded, lefthanded,
189 ; double, triple cutting, 205 ;
sharp " V," 190 ; U.S. Std., 192 ;
Square, 193; Acme 29°, 194;
Whitworth, 196.
Thread cutting, 197, 200.
Toise, 54.
Trapezium, 111.
Triangle, 109, 110.
Trigonometrical ratios, 182.
Try square, 171.
Variation, 288.
Vernier, 173, 174.
Vernier caliper, 177.
W
AVedge, 258.
Whole depth, 208.
Working depth, 208.